YES

The TRS could be proven terminating. The proof took 556 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (18ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (362ms).
 |    | – Problem 4 was processed with processor DependencyGraph (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

g#(s(x))-#(s(x), f(g(x)))f#(s(x))g#(f(x))
g#(s(x))f#(g(x))f#(s(x))-#(s(x), g(f(x)))
-#(s(x), s(y))-#(x, y)f#(s(x))f#(x)
g#(s(x))g#(x)

Rewrite Rules

-(x, 0)x-(0, s(y))0
-(s(x), s(y))-(x, y)f(0)0
f(s(x))-(s(x), g(f(x)))g(0)s(0)
g(s(x))-(s(x), f(g(x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 0, s, -

Strategy


The following SCCs where found

f#(s(x)) → g#(f(x))g#(s(x)) → f#(g(x))
f#(s(x)) → f#(x)g#(s(x)) → g#(x)

-#(s(x), s(y)) → -#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

-#(s(x), s(y))-#(x, y)

Rewrite Rules

-(x, 0)x-(0, s(y))0
-(s(x), s(y))-(x, y)f(0)0
f(s(x))-(s(x), g(f(x)))g(0)s(0)
g(s(x))-(s(x), f(g(x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 0, s, -

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

-#(s(x), s(y))-#(x, y)

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(s(x))g#(f(x))g#(s(x))f#(g(x))
f#(s(x))f#(x)g#(s(x))g#(x)

Rewrite Rules

-(x, 0)x-(0, s(y))0
-(s(x), s(y))-(x, y)f(0)0
f(s(x))-(s(x), g(f(x)))g(0)s(0)
g(s(x))-(s(x), f(g(x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 0, s, -

Strategy


Polynomial Interpretation

Improved Usable rules

-(s(x), s(y))-(x, y)g(0)s(0)
f(s(x))-(s(x), g(f(x)))-(x, 0)x
-(0, s(y))0g(s(x))-(s(x), f(g(x)))
f(0)0

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

g#(s(x))f#(g(x))g#(s(x))g#(x)
f#(s(x))f#(x)

Problem 4: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(s(x))g#(f(x))

Rewrite Rules

-(x, 0)x-(0, s(y))0
-(s(x), s(y))-(x, y)f(0)0
f(s(x))-(s(x), g(f(x)))g(0)s(0)
g(s(x))-(s(x), f(g(x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 0, s, -

Strategy


There are no SCCs!