YES

The TRS could be proven terminating. The proof took 2439 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (207ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (1080ms).
 |    | – Problem 8 was processed with processor PolynomialLinearRange4iUR (850ms).
 |    |    | – Problem 10 was processed with processor DependencyGraph (7ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (95ms).
 |    | – Problem 9 was processed with processor DependencyGraph (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (9ms).
 | – Problem 5 was processed with processor SubtermCriterion (0ms).
 |    | – Problem 7 was processed with processor DependencyGraph (0ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

replace#(N, M, cons(K, L))eq#(N, K)replace#(N, M, cons(K, L))ifrepl#(eq(N, K), N, M, cons(K, L))
ifselsort#(false, cons(N, L))replace#(min(cons(N, L)), N, L)selsort#(cons(N, L))min#(cons(N, L))
ifselsort#(false, cons(N, L))selsort#(replace(min(cons(N, L)), N, L))selsort#(cons(N, L))ifselsort#(eq(N, min(cons(N, L))), cons(N, L))
min#(cons(N, cons(M, L)))ifmin#(le(N, M), cons(N, cons(M, L)))ifselsort#(true, cons(N, L))selsort#(L)
ifmin#(true, cons(N, cons(M, L)))min#(cons(N, L))le#(s(X), s(Y))le#(X, Y)
ifselsort#(false, cons(N, L))min#(cons(N, L))eq#(s(X), s(Y))eq#(X, Y)
selsort#(cons(N, L))eq#(N, min(cons(N, L)))ifmin#(false, cons(N, cons(M, L)))min#(cons(M, L))
ifrepl#(false, N, M, cons(K, L))replace#(N, M, L)min#(cons(N, cons(M, L)))le#(N, M)

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, cons, eq, nil

Strategy


The following SCCs where found

eq#(s(X), s(Y)) → eq#(X, Y)

ifselsort#(false, cons(N, L)) → selsort#(replace(min(cons(N, L)), N, L))selsort#(cons(N, L)) → ifselsort#(eq(N, min(cons(N, L))), cons(N, L))
ifselsort#(true, cons(N, L)) → selsort#(L)

replace#(N, M, cons(K, L)) → ifrepl#(eq(N, K), N, M, cons(K, L))ifrepl#(false, N, M, cons(K, L)) → replace#(N, M, L)

min#(cons(N, cons(M, L))) → ifmin#(le(N, M), cons(N, cons(M, L)))ifmin#(false, cons(N, cons(M, L))) → min#(cons(M, L))
ifmin#(true, cons(N, cons(M, L))) → min#(cons(N, L))

le#(s(X), s(Y)) → le#(X, Y)

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

ifselsort#(false, cons(N, L))selsort#(replace(min(cons(N, L)), N, L))selsort#(cons(N, L))ifselsort#(eq(N, min(cons(N, L))), cons(N, L))
ifselsort#(true, cons(N, L))selsort#(L)

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, cons, eq, nil

Strategy


Polynomial Interpretation

Improved Usable rules

replace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))replace(N, M, nil)nil
ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))ifrepl(true, N, M, cons(K, L))cons(M, L)

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

ifselsort#(true, cons(N, L))selsort#(L)

Problem 8: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

ifselsort#(false, cons(N, L))selsort#(replace(min(cons(N, L)), N, L))selsort#(cons(N, L))ifselsort#(eq(N, min(cons(N, L))), cons(N, L))

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, nil, cons, eq

Strategy


Polynomial Interpretation

Improved Usable rules

replace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))replace(N, M, nil)nil
ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))ifrepl(true, N, M, cons(K, L))cons(M, L)

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

ifselsort#(false, cons(N, L))selsort#(replace(min(cons(N, L)), N, L))

Problem 10: DependencyGraph



Dependency Pair Problem

Dependency Pairs

selsort#(cons(N, L))ifselsort#(eq(N, min(cons(N, L))), cons(N, L))

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, eq, nil, cons

Strategy


There are no SCCs!

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

min#(cons(N, cons(M, L)))ifmin#(le(N, M), cons(N, cons(M, L)))ifmin#(false, cons(N, cons(M, L)))min#(cons(M, L))
ifmin#(true, cons(N, cons(M, L)))min#(cons(N, L))

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, cons, eq, nil

Strategy


Polynomial Interpretation

Improved Usable rules

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

min#(cons(N, cons(M, L)))ifmin#(le(N, M), cons(N, cons(M, L)))

Problem 9: DependencyGraph



Dependency Pair Problem

Dependency Pairs

ifmin#(false, cons(N, cons(M, L)))min#(cons(M, L))ifmin#(true, cons(N, cons(M, L)))min#(cons(N, L))

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, nil, cons, eq

Strategy


There are no SCCs!

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

eq#(s(X), s(Y))eq#(X, Y)

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, cons, eq, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

eq#(s(X), s(Y))eq#(X, Y)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

replace#(N, M, cons(K, L))ifrepl#(eq(N, K), N, M, cons(K, L))ifrepl#(false, N, M, cons(K, L))replace#(N, M, L)

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, cons, eq, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ifrepl#(false, N, M, cons(K, L))replace#(N, M, L)

Problem 7: DependencyGraph



Dependency Pair Problem

Dependency Pairs

replace#(N, M, cons(K, L))ifrepl#(eq(N, K), N, M, cons(K, L))

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, nil, cons, eq

Strategy


There are no SCCs!

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(X), s(Y))le#(X, Y)

Rewrite Rules

eq(0, 0)trueeq(0, s(Y))false
eq(s(X), 0)falseeq(s(X), s(Y))eq(X, Y)
le(0, Y)truele(s(X), 0)false
le(s(X), s(Y))le(X, Y)min(cons(0, nil))0
min(cons(s(N), nil))s(N)min(cons(N, cons(M, L)))ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L)))min(cons(N, L))ifmin(false, cons(N, cons(M, L)))min(cons(M, L))
replace(N, M, nil)nilreplace(N, M, cons(K, L))ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L))cons(M, L)ifrepl(false, N, M, cons(K, L))cons(K, replace(N, M, L))
selsort(nil)nilselsort(cons(N, L))ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L))cons(N, selsort(L))ifselsort(false, cons(N, L))cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Original Signature

Termination of terms over the following signature is verified: min, ifselsort, true, ifmin, replace, ifrepl, 0, s, le, selsort, false, cons, eq, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(X), s(Y))le#(X, Y)