TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60056 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (47ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (752ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (6292ms), DependencyGraph (1ms), ReductionPairSAT (timeout)].
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| cond#(true, x, y, z) | → | cond#(and(gr(x, z), gr(y, z)), p(x), p(y), z) |
Rewrite Rules
| cond(true, x, y, z) | → | cond(and(gr(x, z), gr(y, z)), p(x), p(y), z) | | and(true, true) | → | true |
| and(x, false) | → | false | | and(false, x) | → | false |
| gr(0, 0) | → | false | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, p, false, true, gr, cond, and
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| cond#(true, x, y, z) | → | gr#(x, z) | | cond#(true, x, y, z) | → | p#(y) |
| gr#(s(x), s(y)) | → | gr#(x, y) | | cond#(true, x, y, z) | → | and#(gr(x, z), gr(y, z)) |
| cond#(true, x, y, z) | → | cond#(and(gr(x, z), gr(y, z)), p(x), p(y), z) | | cond#(true, x, y, z) | → | gr#(y, z) |
| cond#(true, x, y, z) | → | p#(x) |
Rewrite Rules
| cond(true, x, y, z) | → | cond(and(gr(x, z), gr(y, z)), p(x), p(y), z) | | and(true, true) | → | true |
| and(x, false) | → | false | | and(false, x) | → | false |
| gr(0, 0) | → | false | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, p, true, false, gr, cond, and
Strategy
The following SCCs where found
| gr#(s(x), s(y)) → gr#(x, y) |
| cond#(true, x, y, z) → cond#(and(gr(x, z), gr(y, z)), p(x), p(y), z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gr#(s(x), s(y)) | → | gr#(x, y) |
Rewrite Rules
| cond(true, x, y, z) | → | cond(and(gr(x, z), gr(y, z)), p(x), p(y), z) | | and(true, true) | → | true |
| and(x, false) | → | false | | and(false, x) | → | false |
| gr(0, 0) | → | false | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, p, true, false, gr, cond, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gr#(s(x), s(y)) | → | gr#(x, y) |