YES

The TRS could be proven terminating. The proof took 331 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (3ms).
 | – Problem 2 was processed with processor BackwardsNarrowing (3ms).
 |    | – Problem 3 was processed with processor ForwardNarrowing (1ms).
 |    | – Problem 4 was processed with processor BackwardsNarrowing (3ms).
 |    |    | – Problem 5 was processed with processor BackwardsNarrowing (3ms).
 |    |    |    | – Problem 6 was processed with processor BackwardsNarrowing (5ms).

Problem 1: DependencyGraph

Dependency Pair Problem

Dependency Pairs

f#(s(0)) → f#(p(s(0)))T(f(s(0))) → f#(s(0))
f#(s(0)) → p#(s(0))

Rewrite Rules

f(0) → cons(0, f(s(0)))f(s(0)) → f(p(s(0)))
p(s(X)) → X

Original Signature

Termination of terms over the following signature is verified: f, 0, s, p, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(f) = μ(f#) = μ(p#) = μ(s) = μ(p) = μ(cons) = {1}

The following SCCs where found

f#(s(0)) → f#(p(s(0)))

Problem 2: BackwardsNarrowing

Dependency Pair Problem

Dependency Pairs

f#(s(0)) → f#(p(s(0)))

Rewrite Rules

f(0) → cons(0, f(s(0)))f(s(0)) → f(p(s(0)))
p(s(X)) → X

Original Signature

Termination of terms over the following signature is verified: f, 0, s, p, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(f) = μ(f#) = μ(p#) = μ(s) = μ(p) = μ(cons) = {1}

The left-hand side of the rule f#(s(0)) → f#(p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
f#(s(p(s(0)))) 
f#(p(s(s(0)))) 
Thus, the rule f#(s(0)) → f#(p(s(0))) is replaced by the following rules:
f#(p(s(s(0)))) → f#(p(s(0)))f#(s(p(s(0)))) → f#(p(s(0)))

Problem 3: ForwardNarrowing

Dependency Pair Problem

Dependency Pairs

f#(s(0)) → f#(0)

Rewrite Rules

f(0) → cons(0, f(s(0)))f(s(0)) → f(p(s(0)))
p(s(X)) → X

Original Signature

Termination of terms over the following signature is verified: f, 0, s, p, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(f) = μ(f#) = μ(p#) = μ(s) = μ(p) = μ(cons) = {1}

The right-hand side of the rule f#(s(0)) → f#(0) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
Thus, the rule f#(s(0)) → f#(0) is deleted.

Problem 4: BackwardsNarrowing

Dependency Pair Problem

Dependency Pairs

f#(p(s(s(0)))) → f#(p(s(0)))f#(s(p(s(0)))) → f#(p(s(0)))

Rewrite Rules

f(0) → cons(0, f(s(0)))f(s(0)) → f(p(s(0)))
p(s(X)) → X

Original Signature

Termination of terms over the following signature is verified: f, 0, s, p, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(f) = μ(f#) = μ(p#) = μ(s) = μ(p) = μ(cons) = {1}

The left-hand side of the rule f#(p(s(s(0)))) → f#(p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
f#(p(p(s(s(s(0)))))) 
f#(p(s(p(s(s(0)))))) 
f#(p(s(s(p(s(0)))))) 
Thus, the rule f#(p(s(s(0)))) → f#(p(s(0))) is replaced by the following rules:
f#(p(s(s(p(s(0)))))) → f#(p(s(0)))f#(p(p(s(s(s(0)))))) → f#(p(s(0)))
f#(p(s(p(s(s(0)))))) → f#(p(s(0)))

Problem 5: BackwardsNarrowing

Dependency Pair Problem

Dependency Pairs

f#(p(s(s(p(s(0)))))) → f#(p(s(0)))f#(p(p(s(s(s(0)))))) → f#(p(s(0)))
f#(p(s(p(s(s(0)))))) → f#(p(s(0)))f#(s(p(s(0)))) → f#(p(s(0)))

Rewrite Rules

f(0) → cons(0, f(s(0)))f(s(0)) → f(p(s(0)))
p(s(X)) → X

Original Signature

Termination of terms over the following signature is verified: f, 0, s, p, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(f) = μ(f#) = μ(p#) = μ(s) = μ(p) = μ(cons) = {1}

The left-hand side of the rule f#(p(s(s(p(s(0)))))) → f#(p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
f#(p(p(s(s(s(p(s(0)))))))) 
f#(p(s(p(s(s(p(s(0)))))))) 
f#(p(s(s(p(p(s(s(0)))))))) 
f#(p(s(s(p(s(p(s(0)))))))) 
Thus, the rule f#(p(s(s(p(s(0)))))) → f#(p(s(0))) is replaced by the following rules:
f#(p(s(s(p(s(p(s(0)))))))) → f#(p(s(0)))f#(p(s(s(p(p(s(s(0)))))))) → f#(p(s(0)))
f#(p(p(s(s(s(p(s(0)))))))) → f#(p(s(0)))f#(p(s(p(s(s(p(s(0)))))))) → f#(p(s(0)))

Problem 6: BackwardsNarrowing

Dependency Pair Problem

Dependency Pairs

f#(p(s(s(p(s(p(s(0)))))))) → f#(p(s(0)))f#(p(p(s(s(s(0)))))) → f#(p(s(0)))
f#(p(s(s(p(p(s(s(0)))))))) → f#(p(s(0)))f#(p(p(s(s(s(p(s(0)))))))) → f#(p(s(0)))
f#(p(s(p(s(s(0)))))) → f#(p(s(0)))f#(p(s(p(s(s(p(s(0)))))))) → f#(p(s(0)))
f#(s(p(s(0)))) → f#(p(s(0)))

Rewrite Rules

f(0) → cons(0, f(s(0)))f(s(0)) → f(p(s(0)))
p(s(X)) → X

Original Signature

Termination of terms over the following signature is verified: f, 0, s, p, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(f) = μ(f#) = μ(p#) = μ(s) = μ(p) = μ(cons) = {1}

The left-hand side of the rule f#(p(s(s(p(s(p(s(0)))))))) → f#(p(s(0))) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
f#(p(s(s(p(p(s(s(p(s(0)))))))))) 
f#(p(s(s(p(s(p(p(s(s(0)))))))))) 
f#(p(s(s(p(s(p(s(p(s(0)))))))))) 
f#(p(s(p(s(s(p(s(p(s(0)))))))))) 
f#(p(p(s(s(s(p(s(p(s(0)))))))))) 
Thus, the rule f#(p(s(s(p(s(p(s(0)))))))) → f#(p(s(0))) is replaced by the following rules:
f#(p(s(s(p(p(s(s(p(s(0)))))))))) → f#(p(s(0)))f#(p(s(p(s(s(p(s(p(s(0)))))))))) → f#(p(s(0)))
f#(p(p(s(s(s(p(s(p(s(0)))))))))) → f#(p(s(0)))f#(p(s(s(p(s(p(s(p(s(0)))))))))) → f#(p(s(0)))
f#(p(s(s(p(s(p(p(s(s(0)))))))))) → f#(p(s(0)))