YES

The TRS could be proven terminating. The proof took 1936 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (54ms).
 | – Problem 2 was processed with processor PolynomialOrderingProcessor (450ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (594ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(x, f(s(s(y)), nil))f#(y, f(s(0), nil))f#(x, f(s(s(y)), f(z, w)))f#(s(x), f(y, f(s(z), w)))
L#(f(s(s(y)), f(z, w)))f#(y, f(s(z), w))f#(x, f(s(s(y)), nil))f#(s(x), f(y, f(s(0), nil)))
L#(f(s(s(y)), f(z, w)))f#(s(0), f(y, f(s(z), w)))f#(x, f(s(s(y)), nil))f#(s(0), nil)
L#(f(s(s(y)), f(z, w)))L#(f(s(0), f(y, f(s(z), w))))L#(f(s(s(y)), f(z, w)))f#(s(z), w)
f#(x, f(s(s(y)), f(z, w)))f#(y, f(s(z), w))f#(x, f(s(s(y)), f(z, w)))f#(s(z), w)

Rewrite Rules

f(x, f(s(s(y)), f(z, w)))f(s(x), f(y, f(s(z), w)))L(f(s(s(y)), f(z, w)))L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil))f(s(x), f(y, f(s(0), nil)))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, L, nil

Strategy


The following SCCs where found

f#(x, f(s(s(y)), nil)) → f#(y, f(s(0), nil))f#(x, f(s(s(y)), f(z, w))) → f#(s(x), f(y, f(s(z), w)))
f#(x, f(s(s(y)), nil)) → f#(s(x), f(y, f(s(0), nil)))f#(x, f(s(s(y)), f(z, w))) → f#(y, f(s(z), w))
f#(x, f(s(s(y)), f(z, w))) → f#(s(z), w)

L#(f(s(s(y)), f(z, w))) → L#(f(s(0), f(y, f(s(z), w))))

Problem 2: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

L#(f(s(s(y)), f(z, w)))L#(f(s(0), f(y, f(s(z), w))))

Rewrite Rules

f(x, f(s(s(y)), f(z, w)))f(s(x), f(y, f(s(z), w)))L(f(s(s(y)), f(z, w)))L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil))f(s(x), f(y, f(s(0), nil)))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, L, nil

Strategy


Polynomial Interpretation

Improved Usable rules

f(x, f(s(s(y)), f(z, w)))f(s(x), f(y, f(s(z), w)))f(x, f(s(s(y)), nil))f(s(x), f(y, f(s(0), nil)))

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

L#(f(s(s(y)), f(z, w)))L#(f(s(0), f(y, f(s(z), w))))

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(x, f(s(s(y)), nil))f#(y, f(s(0), nil))f#(x, f(s(s(y)), f(z, w)))f#(s(x), f(y, f(s(z), w)))
f#(x, f(s(s(y)), nil))f#(s(x), f(y, f(s(0), nil)))f#(x, f(s(s(y)), f(z, w)))f#(y, f(s(z), w))
f#(x, f(s(s(y)), f(z, w)))f#(s(z), w)

Rewrite Rules

f(x, f(s(s(y)), f(z, w)))f(s(x), f(y, f(s(z), w)))L(f(s(s(y)), f(z, w)))L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil))f(s(x), f(y, f(s(0), nil)))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, L, nil

Strategy


Polynomial Interpretation

Improved Usable rules

f(x, f(s(s(y)), f(z, w)))f(s(x), f(y, f(s(z), w)))f(x, f(s(s(y)), nil))f(s(x), f(y, f(s(0), nil)))

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

f#(x, f(s(s(y)), nil))f#(y, f(s(0), nil))f#(x, f(s(s(y)), f(z, w)))f#(s(x), f(y, f(s(z), w)))
f#(x, f(s(s(y)), nil))f#(s(x), f(y, f(s(0), nil)))f#(x, f(s(s(y)), f(z, w)))f#(y, f(s(z), w))
f#(x, f(s(s(y)), f(z, w)))f#(s(z), w)