YES

The TRS could be proven terminating. The proof took 2956 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (912ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (0ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 | – Problem 8 was processed with processor SubtermCriterion (1ms).
 | – Problem 9 was processed with processor SubtermCriterion (0ms).
 |    | – Problem 12 was processed with processor SubtermCriterion (0ms).
 | – Problem 10 was processed with processor SubtermCriterion (0ms).
 | – Problem 11 was processed with processor PolynomialLinearRange4iUR (1273ms).
 |    | – Problem 13 was processed with processor PolynomialLinearRange4iUR (646ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

proper#(cons(X1, X2))proper#(X1)top#(ok(X))top#(active(X))
cons#(ok(X1), ok(X2))cons#(X1, X2)proper#(and(X1, X2))and#(proper(X1), proper(X2))
from#(ok(X))from#(X)active#(first(s(X), cons(Y, Z)))cons#(Y, first(X, Z))
active#(if(X1, X2, X3))active#(X1)active#(add(X1, X2))add#(active(X1), X2)
proper#(and(X1, X2))proper#(X2)first#(mark(X1), X2)first#(X1, X2)
top#(mark(X))proper#(X)proper#(from(X))proper#(X)
proper#(add(X1, X2))proper#(X2)active#(first(X1, X2))active#(X2)
top#(mark(X))top#(proper(X))proper#(cons(X1, X2))proper#(X2)
proper#(first(X1, X2))first#(proper(X1), proper(X2))proper#(add(X1, X2))proper#(X1)
add#(mark(X1), X2)add#(X1, X2)and#(mark(X1), X2)and#(X1, X2)
active#(first(X1, X2))active#(X1)proper#(first(X1, X2))proper#(X2)
active#(from(X))s#(X)proper#(s(X))proper#(X)
active#(add(s(X), Y))add#(X, Y)if#(mark(X1), X2, X3)if#(X1, X2, X3)
proper#(add(X1, X2))add#(proper(X1), proper(X2))active#(first(X1, X2))first#(X1, active(X2))
and#(ok(X1), ok(X2))and#(X1, X2)add#(ok(X1), ok(X2))add#(X1, X2)
proper#(and(X1, X2))proper#(X1)top#(ok(X))active#(X)
active#(first(s(X), cons(Y, Z)))first#(X, Z)active#(and(X1, X2))and#(active(X1), X2)
active#(add(s(X), Y))s#(add(X, Y))active#(from(X))cons#(X, from(s(X)))
proper#(from(X))from#(proper(X))if#(ok(X1), ok(X2), ok(X3))if#(X1, X2, X3)
first#(X1, mark(X2))first#(X1, X2)active#(add(X1, X2))active#(X1)
proper#(if(X1, X2, X3))proper#(X1)proper#(if(X1, X2, X3))proper#(X2)
s#(ok(X))s#(X)first#(ok(X1), ok(X2))first#(X1, X2)
active#(first(X1, X2))first#(active(X1), X2)proper#(cons(X1, X2))cons#(proper(X1), proper(X2))
proper#(if(X1, X2, X3))proper#(X3)proper#(s(X))s#(proper(X))
proper#(first(X1, X2))proper#(X1)active#(if(X1, X2, X3))if#(active(X1), X2, X3)
active#(and(X1, X2))active#(X1)proper#(if(X1, X2, X3))if#(proper(X1), proper(X2), proper(X3))
active#(from(X))from#(s(X))

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


The following SCCs where found

active#(first(X1, X2)) → active#(X2)active#(if(X1, X2, X3)) → active#(X1)
active#(add(X1, X2)) → active#(X1)active#(and(X1, X2)) → active#(X1)
active#(first(X1, X2)) → active#(X1)

add#(mark(X1), X2) → add#(X1, X2)add#(ok(X1), ok(X2)) → add#(X1, X2)

proper#(cons(X1, X2)) → proper#(X1)proper#(if(X1, X2, X3)) → proper#(X1)
proper#(cons(X1, X2)) → proper#(X2)proper#(if(X1, X2, X3)) → proper#(X2)
proper#(add(X1, X2)) → proper#(X1)proper#(and(X1, X2)) → proper#(X1)
proper#(first(X1, X2)) → proper#(X2)proper#(s(X)) → proper#(X)
proper#(and(X1, X2)) → proper#(X2)proper#(first(X1, X2)) → proper#(X1)
proper#(if(X1, X2, X3)) → proper#(X3)proper#(from(X)) → proper#(X)
proper#(add(X1, X2)) → proper#(X2)

cons#(ok(X1), ok(X2)) → cons#(X1, X2)

if#(mark(X1), X2, X3) → if#(X1, X2, X3)if#(ok(X1), ok(X2), ok(X3)) → if#(X1, X2, X3)

s#(ok(X)) → s#(X)

and#(ok(X1), ok(X2)) → and#(X1, X2)and#(mark(X1), X2) → and#(X1, X2)

from#(ok(X)) → from#(X)

top#(mark(X)) → top#(proper(X))top#(ok(X)) → top#(active(X))

first#(ok(X1), ok(X2)) → first#(X1, X2)first#(mark(X1), X2) → first#(X1, X2)
first#(X1, mark(X2)) → first#(X1, X2)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

and#(ok(X1), ok(X2))and#(X1, X2)and#(mark(X1), X2)and#(X1, X2)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

and#(ok(X1), ok(X2))and#(X1, X2)and#(mark(X1), X2)and#(X1, X2)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

s#(ok(X))s#(X)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

s#(ok(X))s#(X)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

add#(mark(X1), X2)add#(X1, X2)add#(ok(X1), ok(X2))add#(X1, X2)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

add#(mark(X1), X2)add#(X1, X2)add#(ok(X1), ok(X2))add#(X1, X2)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

if#(mark(X1), X2, X3)if#(X1, X2, X3)if#(ok(X1), ok(X2), ok(X3))if#(X1, X2, X3)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

if#(mark(X1), X2, X3)if#(X1, X2, X3)if#(ok(X1), ok(X2), ok(X3))if#(X1, X2, X3)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

active#(first(X1, X2))active#(X2)active#(if(X1, X2, X3))active#(X1)
active#(add(X1, X2))active#(X1)active#(and(X1, X2))active#(X1)
active#(first(X1, X2))active#(X1)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

active#(first(X1, X2))active#(X2)active#(if(X1, X2, X3))active#(X1)
active#(add(X1, X2))active#(X1)active#(and(X1, X2))active#(X1)
active#(first(X1, X2))active#(X1)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

proper#(cons(X1, X2))proper#(X1)proper#(if(X1, X2, X3))proper#(X1)
proper#(cons(X1, X2))proper#(X2)proper#(if(X1, X2, X3))proper#(X2)
proper#(add(X1, X2))proper#(X1)proper#(and(X1, X2))proper#(X1)
proper#(first(X1, X2))proper#(X2)proper#(s(X))proper#(X)
proper#(and(X1, X2))proper#(X2)proper#(if(X1, X2, X3))proper#(X3)
proper#(first(X1, X2))proper#(X1)proper#(from(X))proper#(X)
proper#(add(X1, X2))proper#(X2)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

proper#(cons(X1, X2))proper#(X1)proper#(if(X1, X2, X3))proper#(X1)
proper#(cons(X1, X2))proper#(X2)proper#(if(X1, X2, X3))proper#(X2)
proper#(add(X1, X2))proper#(X1)proper#(and(X1, X2))proper#(X1)
proper#(first(X1, X2))proper#(X2)proper#(s(X))proper#(X)
proper#(and(X1, X2))proper#(X2)proper#(if(X1, X2, X3))proper#(X3)
proper#(first(X1, X2))proper#(X1)proper#(from(X))proper#(X)
proper#(add(X1, X2))proper#(X2)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

from#(ok(X))from#(X)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

from#(ok(X))from#(X)

Problem 9: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(ok(X1), ok(X2))first#(X1, X2)first#(mark(X1), X2)first#(X1, X2)
first#(X1, mark(X2))first#(X1, X2)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first#(ok(X1), ok(X2))first#(X1, X2)first#(mark(X1), X2)first#(X1, X2)

Problem 12: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(X1, mark(X2))first#(X1, X2)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, top, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first#(X1, mark(X2))first#(X1, X2)

Problem 10: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons#(ok(X1), ok(X2))cons#(X1, X2)

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

cons#(ok(X1), ok(X2))cons#(X1, X2)

Problem 11: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

top#(mark(X))top#(proper(X))top#(ok(X))top#(active(X))

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, cons, nil, top

Strategy


Polynomial Interpretation

Improved Usable rules

proper(false)ok(false)and(ok(X1), ok(X2))ok(and(X1, X2))
active(if(X1, X2, X3))if(active(X1), X2, X3)active(and(false, Y))mark(false)
first(X1, mark(X2))mark(first(X1, X2))active(first(0, X))mark(nil)
proper(add(X1, X2))add(proper(X1), proper(X2))active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
add(mark(X1), X2)mark(add(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
active(first(X1, X2))first(active(X1), X2)proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))
active(add(X1, X2))add(active(X1), X2)proper(s(X))s(proper(X))
active(and(true, X))mark(X)active(from(X))mark(cons(X, from(s(X))))
first(ok(X1), ok(X2))ok(first(X1, X2))active(and(X1, X2))and(active(X1), X2)
proper(first(X1, X2))first(proper(X1), proper(X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))first(mark(X1), X2)mark(first(X1, X2))
s(ok(X))ok(s(X))active(add(s(X), Y))mark(s(add(X, Y)))
from(ok(X))ok(from(X))cons(ok(X1), ok(X2))ok(cons(X1, X2))
active(if(false, X, Y))mark(Y)proper(true)ok(true)
add(ok(X1), ok(X2))ok(add(X1, X2))proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(nil)ok(nil)active(add(0, X))mark(X)
and(mark(X1), X2)mark(and(X1, X2))proper(0)ok(0)
active(if(true, X, Y))mark(X)proper(from(X))from(proper(X))
active(first(X1, X2))first(X1, active(X2))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

top#(mark(X))top#(proper(X))

Problem 13: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

top#(ok(X))top#(active(X))

Rewrite Rules

active(and(true, X))mark(X)active(and(false, Y))mark(false)
active(if(true, X, Y))mark(X)active(if(false, X, Y))mark(Y)
active(add(0, X))mark(X)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
active(from(X))mark(cons(X, from(s(X))))active(and(X1, X2))and(active(X1), X2)
active(if(X1, X2, X3))if(active(X1), X2, X3)active(add(X1, X2))add(active(X1), X2)
active(first(X1, X2))first(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
and(mark(X1), X2)mark(and(X1, X2))if(mark(X1), X2, X3)mark(if(X1, X2, X3))
add(mark(X1), X2)mark(add(X1, X2))first(mark(X1), X2)mark(first(X1, X2))
first(X1, mark(X2))mark(first(X1, X2))proper(and(X1, X2))and(proper(X1), proper(X2))
proper(true)ok(true)proper(false)ok(false)
proper(if(X1, X2, X3))if(proper(X1), proper(X2), proper(X3))proper(add(X1, X2))add(proper(X1), proper(X2))
proper(0)ok(0)proper(s(X))s(proper(X))
proper(first(X1, X2))first(proper(X1), proper(X2))proper(nil)ok(nil)
proper(cons(X1, X2))cons(proper(X1), proper(X2))proper(from(X))from(proper(X))
and(ok(X1), ok(X2))ok(and(X1, X2))if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))
add(ok(X1), ok(X2))ok(add(X1, X2))s(ok(X))ok(s(X))
first(ok(X1), ok(X2))ok(first(X1, X2))cons(ok(X1), ok(X2))ok(cons(X1, X2))
from(ok(X))ok(from(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: true, mark, from, add, and, 0, s, if, active, false, ok, proper, first, top, nil, cons

Strategy


Polynomial Interpretation

Improved Usable rules

if(ok(X1), ok(X2), ok(X3))ok(if(X1, X2, X3))and(ok(X1), ok(X2))ok(and(X1, X2))
first(mark(X1), X2)mark(first(X1, X2))active(if(X1, X2, X3))if(active(X1), X2, X3)
active(and(false, Y))mark(false)active(add(s(X), Y))mark(s(add(X, Y)))
active(first(0, X))mark(nil)first(X1, mark(X2))mark(first(X1, X2))
active(if(false, X, Y))mark(Y)active(first(s(X), cons(Y, Z)))mark(cons(Y, first(X, Z)))
add(ok(X1), ok(X2))ok(add(X1, X2))add(mark(X1), X2)mark(add(X1, X2))
active(first(X1, X2))first(active(X1), X2)active(add(X1, X2))add(active(X1), X2)
active(add(0, X))mark(X)and(mark(X1), X2)mark(and(X1, X2))
active(and(true, X))mark(X)active(from(X))mark(cons(X, from(s(X))))
first(ok(X1), ok(X2))ok(first(X1, X2))active(if(true, X, Y))mark(X)
active(and(X1, X2))and(active(X1), X2)active(first(X1, X2))first(X1, active(X2))
if(mark(X1), X2, X3)mark(if(X1, X2, X3))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

top#(ok(X))top#(active(X))