YES

The TRS could be proven terminating. The proof took 1157 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (87ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (89ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4 (37ms).
 | – Problem 4 was processed with processor PolynomialLinearRange4 (292ms).
 |    | – Problem 5 was processed with processor DependencyGraph (5ms).
 |    |    | – Problem 6 was processed with processor PolynomialLinearRange4 (137ms).
 |    |    |    | – Problem 7 was processed with processor PolynomialLinearRange4 (72ms).
 |    |    |    |    | – Problem 8 was processed with processor PolynomialLinearRange4 (15ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(false, X, Y)T(Y)T(minus(x_1, x_2))T(x_2)
T(div(minus(X, Y), s(Y)))div#(minus(X, Y), s(Y))T(div(x_1, x_2))T(x_2)
T(minus(x_1, x_2))T(x_1)div#(s(X), s(Y))geq#(X, Y)
div#(s(X), s(Y))if#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)T(div(x_1, x_2))T(x_1)
geq#(s(X), s(Y))geq#(X, Y)minus#(s(X), s(Y))minus#(X, Y)
T(s(x_1))T(x_1)T(minus(X, Y))minus#(X, Y)
if#(true, X, Y)T(X)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


The following SCCs where found

geq#(s(X), s(Y)) → geq#(X, Y)

minus#(s(X), s(Y)) → minus#(X, Y)

div#(s(X), s(Y)) → if#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)T(div(x_1, x_2)) → T(x_1)
T(minus(x_1, x_2)) → T(x_2)if#(false, X, Y) → T(Y)
T(div(minus(X, Y), s(Y))) → div#(minus(X, Y), s(Y))T(s(x_1)) → T(x_1)
T(div(x_1, x_2)) → T(x_2)T(minus(x_1, x_2)) → T(x_1)
if#(true, X, Y) → T(X)

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

minus#(s(X), s(Y))minus#(X, Y)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


Polynomial Interpretation

There are no usable rules

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

minus#(s(X), s(Y))minus#(X, Y)

Problem 3: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

geq#(s(X), s(Y))geq#(X, Y)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


Polynomial Interpretation

There are no usable rules

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

geq#(s(X), s(Y))geq#(X, Y)

Problem 4: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

div#(s(X), s(Y))if#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)T(div(x_1, x_2))T(x_1)
T(minus(x_1, x_2))T(x_2)if#(false, X, Y)T(Y)
T(div(minus(X, Y), s(Y)))div#(minus(X, Y), s(Y))T(s(x_1))T(x_1)
T(div(x_1, x_2))T(x_2)T(minus(x_1, x_2))T(x_1)
if#(true, X, Y)T(X)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


Polynomial Interpretation

Standard Usable rules

geq(0, s(Y))falseif(false, X, Y)Y
if(true, X, Y)Xdiv(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
minus(0, Y)0geq(X, 0)true
geq(s(X), s(Y))geq(X, Y)minus(s(X), s(Y))minus(X, Y)
div(0, s(Y))0

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

div#(s(X), s(Y))if#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)

Problem 5: DependencyGraph



Dependency Pair Problem

Dependency Pairs

T(div(x_1, x_2))T(x_1)if#(false, X, Y)T(Y)
T(minus(x_1, x_2))T(x_2)T(div(minus(X, Y), s(Y)))div#(minus(X, Y), s(Y))
T(s(x_1))T(x_1)T(div(x_1, x_2))T(x_2)
T(minus(x_1, x_2))T(x_1)if#(true, X, Y)T(X)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, 0, minus, s, if, div, false, true

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


The following SCCs where found

T(div(x_1, x_2)) → T(x_1)T(minus(x_1, x_2)) → T(x_2)
T(s(x_1)) → T(x_1)T(div(x_1, x_2)) → T(x_2)
T(minus(x_1, x_2)) → T(x_1)

Problem 6: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(div(x_1, x_2))T(x_1)T(minus(x_1, x_2))T(x_2)
T(s(x_1))T(x_1)T(div(x_1, x_2))T(x_2)
T(minus(x_1, x_2))T(x_1)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, 0, minus, s, if, div, false, true

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


Polynomial Interpretation

There are no usable rules

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

T(s(x_1))T(x_1)

Problem 7: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(div(x_1, x_2))T(x_1)T(minus(x_1, x_2))T(x_2)
T(div(x_1, x_2))T(x_2)T(minus(x_1, x_2))T(x_1)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, minus, 0, s, if, div, true, false

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


Polynomial Interpretation

There are no usable rules

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

T(minus(x_1, x_2))T(x_2)T(minus(x_1, x_2))T(x_1)

Problem 8: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(div(x_1, x_2))T(x_1)T(div(x_1, x_2))T(x_2)

Rewrite Rules

minus(0, Y)0minus(s(X), s(Y))minus(X, Y)
geq(X, 0)truegeq(0, s(Y))false
geq(s(X), s(Y))geq(X, Y)div(0, s(Y))0
div(s(X), s(Y))if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)if(true, X, Y)X
if(false, X, Y)Y

Original Signature

Termination of terms over the following signature is verified: geq, 0, minus, s, if, div, false, true

Strategy

Context-sensitive strategy:
μ(geq) = μ(minus) = μ(geq#) = μ(true) = μ(minus#) = μ(T) = μ(0) = μ(false) = ∅
μ(div) = μ(div#) = μ(s) = μ(if) = μ(if#) = {1}


Polynomial Interpretation

There are no usable rules

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

T(div(x_1, x_2))T(x_1)T(div(x_1, x_2))T(x_2)