TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60000 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (103ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (11ms), PolynomialLinearRange4iUR (995ms), DependencyGraph (8ms), PolynomialLinearRange8NegiUR (12834ms), DependencyGraph (7ms), ReductionPairSAT (timeout)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| cond3#(true, x, y, z) | → | cond1#(gr(add(x, y), z), x, p(y), z) | | cond2#(false, x, y, z) | → | cond3#(gr(y, 0), x, y, z) |
| cond2#(true, x, y, z) | → | cond1#(gr(add(x, y), z), p(x), y, z) | | cond3#(false, x, y, z) | → | cond1#(gr(add(x, y), z), x, y, z) |
| cond1#(true, x, y, z) | → | cond2#(gr(x, 0), x, y, z) |
Rewrite Rules
| cond1(true, x, y, z) | → | cond2(gr(x, 0), x, y, z) | | cond2(true, x, y, z) | → | cond1(gr(add(x, y), z), p(x), y, z) |
| cond2(false, x, y, z) | → | cond3(gr(y, 0), x, y, z) | | cond3(true, x, y, z) | → | cond1(gr(add(x, y), z), x, p(y), z) |
| cond3(false, x, y, z) | → | cond1(gr(add(x, y), z), x, y, z) | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| add(0, x) | → | x | | add(s(x), y) | → | s(add(x, y)) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: cond2, cond3, 0, s, p, false, true, add, gr, cond1
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| cond2#(false, x, y, z) | → | gr#(y, 0) | | cond3#(true, x, y, z) | → | cond1#(gr(add(x, y), z), x, p(y), z) |
| cond2#(true, x, y, z) | → | p#(x) | | cond2#(true, x, y, z) | → | gr#(add(x, y), z) |
| cond3#(true, x, y, z) | → | add#(x, y) | | cond3#(true, x, y, z) | → | p#(y) |
| cond2#(true, x, y, z) | → | add#(x, y) | | cond3#(false, x, y, z) | → | gr#(add(x, y), z) |
| cond3#(true, x, y, z) | → | gr#(add(x, y), z) | | cond3#(false, x, y, z) | → | add#(x, y) |
| cond2#(false, x, y, z) | → | cond3#(gr(y, 0), x, y, z) | | cond2#(true, x, y, z) | → | cond1#(gr(add(x, y), z), p(x), y, z) |
| cond1#(true, x, y, z) | → | gr#(x, 0) | | gr#(s(x), s(y)) | → | gr#(x, y) |
| add#(s(x), y) | → | add#(x, y) | | cond3#(false, x, y, z) | → | cond1#(gr(add(x, y), z), x, y, z) |
| cond1#(true, x, y, z) | → | cond2#(gr(x, 0), x, y, z) |
Rewrite Rules
| cond1(true, x, y, z) | → | cond2(gr(x, 0), x, y, z) | | cond2(true, x, y, z) | → | cond1(gr(add(x, y), z), p(x), y, z) |
| cond2(false, x, y, z) | → | cond3(gr(y, 0), x, y, z) | | cond3(true, x, y, z) | → | cond1(gr(add(x, y), z), x, p(y), z) |
| cond3(false, x, y, z) | → | cond1(gr(add(x, y), z), x, y, z) | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| add(0, x) | → | x | | add(s(x), y) | → | s(add(x, y)) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: cond2, cond3, 0, s, p, true, false, gr, add, cond1
Strategy
The following SCCs where found
| gr#(s(x), s(y)) → gr#(x, y) |
| cond2#(false, x, y, z) → cond3#(gr(y, 0), x, y, z) | cond3#(true, x, y, z) → cond1#(gr(add(x, y), z), x, p(y), z) |
| cond2#(true, x, y, z) → cond1#(gr(add(x, y), z), p(x), y, z) | cond3#(false, x, y, z) → cond1#(gr(add(x, y), z), x, y, z) |
| cond1#(true, x, y, z) → cond2#(gr(x, 0), x, y, z) |
| add#(s(x), y) → add#(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(x, 0), x, y, z) | | cond2(true, x, y, z) | → | cond1(gr(add(x, y), z), p(x), y, z) |
| cond2(false, x, y, z) | → | cond3(gr(y, 0), x, y, z) | | cond3(true, x, y, z) | → | cond1(gr(add(x, y), z), x, p(y), z) |
| cond3(false, x, y, z) | → | cond1(gr(add(x, y), z), x, y, z) | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| add(0, x) | → | x | | add(s(x), y) | → | s(add(x, y)) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: cond2, cond3, 0, s, p, true, false, gr, add, cond1
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gr#(s(x), s(y)) | → | gr#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(x), y) | → | add#(x, y) |
Rewrite Rules
| cond1(true, x, y, z) | → | cond2(gr(x, 0), x, y, z) | | cond2(true, x, y, z) | → | cond1(gr(add(x, y), z), p(x), y, z) |
| cond2(false, x, y, z) | → | cond3(gr(y, 0), x, y, z) | | cond3(true, x, y, z) | → | cond1(gr(add(x, y), z), x, p(y), z) |
| cond3(false, x, y, z) | → | cond1(gr(add(x, y), z), x, y, z) | | gr(0, x) | → | false |
| gr(s(x), 0) | → | true | | gr(s(x), s(y)) | → | gr(x, y) |
| add(0, x) | → | x | | add(s(x), y) | → | s(add(x, y)) |
| p(0) | → | 0 | | p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: cond2, cond3, 0, s, p, true, false, gr, add, cond1
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(x), y) | → | add#(x, y) |