YES

The TRS could be proven terminating. The proof took 1696 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (60ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor PolynomialOrderingProcessor (170ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor PolynomialOrderingProcessor (649ms).
 | – Problem 6 was processed with processor PolynomialOrderingProcessor (119ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

towerIter#(s(x), y, z)towerIter#(p(s(x)), y, exp(y, z))times#(s(x), y)plus#(y, times(p(s(x)), y))
exp#(x, s(y))exp#(x, y)plus#(s(x), y)plus#(p(s(x)), y)
times#(s(x), y)p#(s(x))tower#(x, y)towerIter#(x, y, s(0))
towerIter#(s(x), y, z)p#(s(x))towerIter#(s(x), y, z)exp#(y, z)
times#(s(x), y)times#(p(s(x)), y)plus#(s(x), y)p#(s(x))
exp#(x, s(y))times#(x, exp(x, y))p#(s(s(x)))p#(s(x))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(p(s(x)), y))
times(0, y)0times(s(x), y)plus(y, times(p(s(x)), y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
p(s(0))0p(s(s(x)))s(p(s(x)))
tower(x, y)towerIter(x, y, s(0))towerIter(0, y, z)z
towerIter(s(x), y, z)towerIter(p(s(x)), y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, 0, s, times, p, towerIter

Strategy


The following SCCs where found

towerIter#(s(x), y, z) → towerIter#(p(s(x)), y, exp(y, z))

exp#(x, s(y)) → exp#(x, y)

plus#(s(x), y) → plus#(p(s(x)), y)

times#(s(x), y) → times#(p(s(x)), y)

p#(s(s(x))) → p#(s(x))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

exp#(x, s(y))exp#(x, y)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(p(s(x)), y))
times(0, y)0times(s(x), y)plus(y, times(p(s(x)), y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
p(s(0))0p(s(s(x)))s(p(s(x)))
tower(x, y)towerIter(x, y, s(0))towerIter(0, y, z)z
towerIter(s(x), y, z)towerIter(p(s(x)), y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, 0, s, times, p, towerIter

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

exp#(x, s(y))exp#(x, y)

Problem 3: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(p(s(x)), y))
times(0, y)0times(s(x), y)plus(y, times(p(s(x)), y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
p(s(0))0p(s(s(x)))s(p(s(x)))
tower(x, y)towerIter(x, y, s(0))towerIter(0, y, z)z
towerIter(s(x), y, z)towerIter(p(s(x)), y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, 0, s, times, p, towerIter

Strategy


Polynomial Interpretation

Improved Usable rules

p(s(s(x)))s(p(s(x)))p(s(0))0

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

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

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

p#(s(s(x)))p#(s(x))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(p(s(x)), y))
times(0, y)0times(s(x), y)plus(y, times(p(s(x)), y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
p(s(0))0p(s(s(x)))s(p(s(x)))
tower(x, y)towerIter(x, y, s(0))towerIter(0, y, z)z
towerIter(s(x), y, z)towerIter(p(s(x)), y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, 0, s, times, p, towerIter

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

p#(s(s(x)))p#(s(x))

Problem 5: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

towerIter#(s(x), y, z)towerIter#(p(s(x)), y, exp(y, z))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(p(s(x)), y))
times(0, y)0times(s(x), y)plus(y, times(p(s(x)), y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
p(s(0))0p(s(s(x)))s(p(s(x)))
tower(x, y)towerIter(x, y, s(0))towerIter(0, y, z)z
towerIter(s(x), y, z)towerIter(p(s(x)), y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, 0, s, times, p, towerIter

Strategy


Polynomial Interpretation

Improved Usable rules

p(s(s(x)))s(p(s(x)))p(s(0))0

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

towerIter#(s(x), y, z)towerIter#(p(s(x)), y, exp(y, z))

Problem 6: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(p(s(x)), y))
times(0, y)0times(s(x), y)plus(y, times(p(s(x)), y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
p(s(0))0p(s(s(x)))s(p(s(x)))
tower(x, y)towerIter(x, y, s(0))towerIter(0, y, z)z
towerIter(s(x), y, z)towerIter(p(s(x)), y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, 0, s, times, p, towerIter

Strategy


Polynomial Interpretation

Improved Usable rules

p(s(s(x)))s(p(s(x)))p(s(0))0

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

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