YES

The TRS could be proven terminating. The proof took 36 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (16ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

f(0)1f(s(x))g(x, s(x))
g(0, y)yg(s(x), y)g(x, +(y, s(x)))
+(x, 0)x+(x, s(y))s(+(x, y))
g(s(x), y)g(x, s(+(y, x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, +

Strategy


The following SCCs where found

g#(s(x), y) → g#(x, +(y, s(x)))g#(s(x), y) → g#(x, s(+(y, x)))

+#(x, s(y)) → +#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

g#(s(x), y)g#(x, +(y, s(x)))g#(s(x), y)g#(x, s(+(y, x)))

Rewrite Rules

f(0)1f(s(x))g(x, s(x))
g(0, y)yg(s(x), y)g(x, +(y, s(x)))
+(x, 0)x+(x, s(y))s(+(x, y))
g(s(x), y)g(x, s(+(y, x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, +

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

g#(s(x), y)g#(x, +(y, s(x)))g#(s(x), y)g#(x, s(+(y, x)))

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

+#(x, s(y))+#(x, y)

Rewrite Rules

f(0)1f(s(x))g(x, s(x))
g(0, y)yg(s(x), y)g(x, +(y, s(x)))
+(x, 0)x+(x, s(y))s(+(x, y))
g(s(x), y)g(x, s(+(y, x)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, +

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

+#(x, s(y))+#(x, y)