YES

The TRS could be proven terminating. The proof took 307 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (52ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (2ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (151ms).
 |    |    | – Problem 6 was processed with processor DependencyGraph (13ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

times#(s(x), y)plus#(y, times(x, y))times#(s(x), y)times#(x, y)
plus#(s(x), y)plus#(x, y)quot#(x, 0, s(z))div#(x, s(z))
div#(div(x, y), z)times#(y, z)div#(x, y)quot#(x, y, y)
div#(div(x, y), z)div#(x, times(y, z))quot#(s(x), s(y), z)quot#(x, y, z)

Rewrite Rules

plus(x, 0)xplus(0, y)y
plus(s(x), y)s(plus(x, y))times(0, y)0
times(s(0), y)ytimes(s(x), y)plus(y, times(x, y))
div(0, y)0div(x, y)quot(x, y, y)
quot(0, s(y), z)0quot(s(x), s(y), z)quot(x, y, z)
quot(x, 0, s(z))s(div(x, s(z)))div(div(x, y), z)div(x, times(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times, div, quot

Strategy


The following SCCs where found

times#(s(x), y) → times#(x, y)

plus#(s(x), y) → plus#(x, y)

quot#(x, 0, s(z)) → div#(x, s(z))div#(x, y) → quot#(x, y, y)
div#(div(x, y), z) → div#(x, times(y, z))quot#(s(x), s(y), z) → quot#(x, y, z)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

times#(s(x), y)times#(x, y)

Rewrite Rules

plus(x, 0)xplus(0, y)y
plus(s(x), y)s(plus(x, y))times(0, y)0
times(s(0), y)ytimes(s(x), y)plus(y, times(x, y))
div(0, y)0div(x, y)quot(x, y, y)
quot(0, s(y), z)0quot(s(x), s(y), z)quot(x, y, z)
quot(x, 0, s(z))s(div(x, s(z)))div(div(x, y), z)div(x, times(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times, div, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

times#(s(x), y)times#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(s(x), y)plus#(x, y)

Rewrite Rules

plus(x, 0)xplus(0, y)y
plus(s(x), y)s(plus(x, y))times(0, y)0
times(s(0), y)ytimes(s(x), y)plus(y, times(x, y))
div(0, y)0div(x, y)quot(x, y, y)
quot(0, s(y), z)0quot(s(x), s(y), z)quot(x, y, z)
quot(x, 0, s(z))s(div(x, s(z)))div(div(x, y), z)div(x, times(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times, div, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

plus#(s(x), y)plus#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

quot#(x, 0, s(z))div#(x, s(z))div#(x, y)quot#(x, y, y)
div#(div(x, y), z)div#(x, times(y, z))quot#(s(x), s(y), z)quot#(x, y, z)

Rewrite Rules

plus(x, 0)xplus(0, y)y
plus(s(x), y)s(plus(x, y))times(0, y)0
times(s(0), y)ytimes(s(x), y)plus(y, times(x, y))
div(0, y)0div(x, y)quot(x, y, y)
quot(0, s(y), z)0quot(s(x), s(y), z)quot(x, y, z)
quot(x, 0, s(z))s(div(x, s(z)))div(div(x, y), z)div(x, times(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times, div, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

div#(div(x, y), z)div#(x, times(y, z))quot#(s(x), s(y), z)quot#(x, y, z)

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

quot#(x, 0, s(z))div#(x, s(z))div#(x, y)quot#(x, y, y)

Rewrite Rules

plus(x, 0)xplus(0, y)y
plus(s(x), y)s(plus(x, y))times(0, y)0
times(s(0), y)ytimes(s(x), y)plus(y, times(x, y))
div(0, y)0div(x, y)quot(x, y, y)
quot(0, s(y), z)0quot(s(x), s(y), z)quot(x, y, z)
quot(x, 0, s(z))s(div(x, s(z)))div(div(x, y), z)div(x, times(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times, div, quot

Strategy


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:

quot#(x, 0, s(z))div#(x, s(z))

Problem 6: DependencyGraph



Dependency Pair Problem

Dependency Pairs

div#(x, y)quot#(x, y, y)

Rewrite Rules

plus(x, 0)xplus(0, y)y
plus(s(x), y)s(plus(x, y))times(0, y)0
times(s(0), y)ytimes(s(x), y)plus(y, times(x, y))
div(0, y)0div(x, y)quot(x, y, y)
quot(0, s(y), z)0quot(s(x), s(y), z)quot(x, y, z)
quot(x, 0, s(z))s(div(x, s(z)))div(div(x, y), z)div(x, times(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times, div, quot

Strategy


There are no SCCs!