YES

The TRS could be proven terminating. The proof took 27 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (13ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(f(x))f#(x)g#(c(h(0)))g#(d(1))
f#(f(x))f#(d(f(x)))g#(c(1))g#(d(h(0)))
g#(h(x))g#(x)f#(f(x))f#(c(f(x)))

Rewrite Rules

f(f(x))f(c(f(x)))f(f(x))f(d(f(x)))
g(c(x))xg(d(x))x
g(c(h(0)))g(d(1))g(c(1))g(d(h(0)))
g(h(x))g(x)

Original Signature

Termination of terms over the following signature is verified: f, g, d, 1, 0, c, h

Strategy


The following SCCs where found

f#(f(x)) → f#(x)

g#(h(x)) → g#(x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(f(x))f#(x)

Rewrite Rules

f(f(x))f(c(f(x)))f(f(x))f(d(f(x)))
g(c(x))xg(d(x))x
g(c(h(0)))g(d(1))g(c(1))g(d(h(0)))
g(h(x))g(x)

Original Signature

Termination of terms over the following signature is verified: f, g, d, 1, 0, c, h

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(f(x))f#(x)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

g#(h(x))g#(x)

Rewrite Rules

f(f(x))f(c(f(x)))f(f(x))f(d(f(x)))
g(c(x))xg(d(x))x
g(c(h(0)))g(d(1))g(c(1))g(d(h(0)))
g(h(x))g(x)

Original Signature

Termination of terms over the following signature is verified: f, g, d, 1, 0, c, h

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

g#(h(x))g#(x)