TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60006 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (47ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (378ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (2490ms), DependencyGraph (4ms), ReductionPairSAT (timeout)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| cond2#(false, x, y, z) | → | cond1#(gr(x, z), p(x), y, z) | | cond1#(true, x, y, z) | → | cond2#(gr(y, z), x, y, z) |
| cond2#(true, x, y, z) | → | cond2#(gr(y, z), x, p(y), z) |
Rewrite Rules
| cond1(true, x, y, z) | → | cond2(gr(y, z), x, y, z) | | cond2(true, x, y, z) | → | cond2(gr(y, z), x, p(y), z) |
| cond2(false, x, y, z) | → | cond1(gr(x, z), p(x), y, z) | | 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: cond2, 0, s, p, false, true, gr, cond1
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| cond2#(false, x, y, z) | → | gr#(x, z) | | cond2#(false, x, y, z) | → | p#(x) |
| cond2#(true, x, y, z) | → | p#(y) | | cond2#(false, x, y, z) | → | cond1#(gr(x, z), p(x), y, z) |
| cond1#(true, x, y, z) | → | gr#(y, z) | | gr#(s(x), s(y)) | → | gr#(x, y) |
| cond2#(true, x, y, z) | → | gr#(y, z) | | cond1#(true, x, y, z) | → | cond2#(gr(y, z), x, y, z) |
| cond2#(true, x, y, z) | → | cond2#(gr(y, z), x, p(y), z) |
Rewrite Rules
| cond1(true, x, y, z) | → | cond2(gr(y, z), x, y, z) | | cond2(true, x, y, z) | → | cond2(gr(y, z), x, p(y), z) |
| cond2(false, x, y, z) | → | cond1(gr(x, z), p(x), y, z) | | 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: cond2, 0, s, p, true, false, gr, cond1
Strategy
The following SCCs where found
| cond2#(false, x, y, z) → cond1#(gr(x, z), p(x), y, z) | cond1#(true, x, y, z) → cond2#(gr(y, z), x, y, z) |
| cond2#(true, x, y, z) → cond2#(gr(y, z), x, p(y), z) |
| gr#(s(x), s(y)) → gr#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gr#(s(x), s(y)) | → | gr#(x, y) |
Rewrite Rules
| cond1(true, x, y, z) | → | cond2(gr(y, z), x, y, z) | | cond2(true, x, y, z) | → | cond2(gr(y, z), x, p(y), z) |
| cond2(false, x, y, z) | → | cond1(gr(x, z), p(x), y, z) | | 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: cond2, 0, s, p, true, false, gr, cond1
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gr#(s(x), s(y)) | → | gr#(x, y) |