YES
The TRS could be proven terminating. The proof took 86 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (67ms).
| – 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#(s(s(x))) | → | p#(h(g(x))) | | g#(s(x)) | → | h#(g(x)) |
| f#(s(s(x))) | → | +#(p(g(x)), q(g(x))) | | f#(s(s(x))) | → | p#(g(x)) |
| f#(s(s(x))) | → | h#(g(x)) | | g#(s(x)) | → | q#(g(x)) |
| g#(s(x)) | → | g#(x) | | h#(x) | → | p#(x) |
| h#(x) | → | +#(p(x), q(x)) | | f#(s(s(x))) | → | q#(g(x)) |
| h#(x) | → | q#(x) | | +#(x, s(y)) | → | +#(x, y) |
| g#(s(x)) | → | +#(p(g(x)), q(g(x))) | | g#(s(x)) | → | p#(g(x)) |
| f#(s(s(x))) | → | g#(x) |
Rewrite Rules
| f(0) | → | 0 | | f(s(0)) | → | s(0) |
| f(s(s(x))) | → | p(h(g(x))) | | g(0) | → | pair(s(0), s(0)) |
| g(s(x)) | → | h(g(x)) | | h(x) | → | pair(+(p(x), q(x)), p(x)) |
| p(pair(x, y)) | → | x | | q(pair(x, y)) | → | y |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| f(s(s(x))) | → | +(p(g(x)), q(g(x))) | | g(s(x)) | → | pair(+(p(g(x)), q(g(x))), p(g(x))) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, pair, q, p, +, h
Strategy
The following SCCs where found
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(0) | → | 0 | | f(s(0)) | → | s(0) |
| f(s(s(x))) | → | p(h(g(x))) | | g(0) | → | pair(s(0), s(0)) |
| g(s(x)) | → | h(g(x)) | | h(x) | → | pair(+(p(x), q(x)), p(x)) |
| p(pair(x, y)) | → | x | | q(pair(x, y)) | → | y |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| f(s(s(x))) | → | +(p(g(x)), q(g(x))) | | g(s(x)) | → | pair(+(p(g(x)), q(g(x))), p(g(x))) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, pair, q, p, +, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(0) | → | 0 | | f(s(0)) | → | s(0) |
| f(s(s(x))) | → | p(h(g(x))) | | g(0) | → | pair(s(0), s(0)) |
| g(s(x)) | → | h(g(x)) | | h(x) | → | pair(+(p(x), q(x)), p(x)) |
| p(pair(x, y)) | → | x | | q(pair(x, y)) | → | y |
| +(x, 0) | → | x | | +(x, s(y)) | → | s(+(x, y)) |
| f(s(s(x))) | → | +(p(g(x)), q(g(x))) | | g(s(x)) | → | pair(+(p(g(x)), q(g(x))), p(g(x))) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s, pair, q, p, +, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: