YES
The TRS could be proven terminating. The proof took 24 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (11ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y), z, u) | → | f#(s(x), minus(y, x), z, u) | | f#(s(x), 0, z, u) | → | f#(x, u, minus(z, s(x)), u) |
| f#(s(x), s(y), z, u) | → | f#(x, u, z, u) | | perfectp#(s(x)) | → | f#(x, s(0), s(x), s(x)) |
Rewrite Rules
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, 0, minus, s, le, if, perfectp, false, true
Strategy
The following SCCs where found
| f#(s(x), 0, z, u) → f#(x, u, minus(z, s(x)), u) | f#(s(x), s(y), z, u) → f#(x, u, z, u) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(s(x), 0, z, u) | → | f#(x, u, minus(z, s(x)), u) | | f#(s(x), s(y), z, u) | → | f#(x, u, z, u) |
Rewrite Rules
| perfectp(0) | → | false | | perfectp(s(x)) | → | f(x, s(0), s(x), s(x)) |
| f(0, y, 0, u) | → | true | | f(0, y, s(z), u) | → | false |
| f(s(x), 0, z, u) | → | f(x, u, minus(z, s(x)), u) | | f(s(x), s(y), z, u) | → | if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) |
Original Signature
Termination of terms over the following signature is verified: f, 0, minus, s, le, if, perfectp, false, true
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(x), s(y), z, u) | → | f#(x, u, z, u) | | f#(s(x), 0, z, u) | → | f#(x, u, minus(z, s(x)), u) |