YES

The TRS could be proven terminating. The proof took 221 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (19ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor PolynomialLinearRange4iUR (156ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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


The following SCCs where found

minus#(s(X), s(Y)) → minus#(X, Y)

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

geq#(s(X), s(Y)) → geq#(X, Y)

Problem 2: SubtermCriterion



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


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: SubtermCriterion



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


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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


Polynomial Interpretation

Improved Usable rules

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