TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60029 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (43ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (10ms), PolynomialLinearRange4iUR (496ms), DependencyGraph (7ms), PolynomialLinearRange8NegiUR (8188ms), DependencyGraph (46ms), ReductionPairSAT (timeout)].
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| if2#(true, x, y, z, u, v) | → | quotIter#(x, y, z, 0, s(v)) | | if2#(false, x, y, z, u, v) | → | quotIter#(x, y, z, u, v) |
| if#(false, x, y, z, u, v) | → | if2#(le(y, s(u)), x, y, s(z), s(u), v) | | quotIter#(x, s(y), z, u, v) | → | if#(le(x, z), x, s(y), z, u, v) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(x, 0) | → | quotZeroErro |
| quot(x, s(y)) | → | quotIter(x, s(y), 0, 0, 0) | | quotIter(x, s(y), z, u, v) | → | if(le(x, z), x, s(y), z, u, v) |
| if(true, x, y, z, u, v) | → | v | | if(false, x, y, z, u, v) | → | if2(le(y, s(u)), x, y, s(z), s(u), v) |
| if2(false, x, y, z, u, v) | → | quotIter(x, y, z, u, v) | | if2(true, x, y, z, u, v) | → | quotIter(x, y, z, 0, s(v)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, le, quotZeroErro, if, quotIter, false, true, if2, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) | | if2#(true, x, y, z, u, v) | → | quotIter#(x, y, z, 0, s(v)) |
| quot#(x, s(y)) | → | quotIter#(x, s(y), 0, 0, 0) | | if2#(false, x, y, z, u, v) | → | quotIter#(x, y, z, u, v) |
| if#(false, x, y, z, u, v) | → | if2#(le(y, s(u)), x, y, s(z), s(u), v) | | if#(false, x, y, z, u, v) | → | le#(y, s(u)) |
| quotIter#(x, s(y), z, u, v) | → | le#(x, z) | | quotIter#(x, s(y), z, u, v) | → | if#(le(x, z), x, s(y), z, u, v) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(x, 0) | → | quotZeroErro |
| quot(x, s(y)) | → | quotIter(x, s(y), 0, 0, 0) | | quotIter(x, s(y), z, u, v) | → | if(le(x, z), x, s(y), z, u, v) |
| if(true, x, y, z, u, v) | → | v | | if(false, x, y, z, u, v) | → | if2(le(y, s(u)), x, y, s(z), s(u), v) |
| if2(false, x, y, z, u, v) | → | quotIter(x, y, z, u, v) | | if2(true, x, y, z, u, v) | → | quotIter(x, y, z, 0, s(v)) |
Original Signature
Termination of terms over the following signature is verified: 0, le, s, quotZeroErro, quotIter, if, true, false, if2, quot
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| if2#(true, x, y, z, u, v) → quotIter#(x, y, z, 0, s(v)) | if2#(false, x, y, z, u, v) → quotIter#(x, y, z, u, v) |
| if#(false, x, y, z, u, v) → if2#(le(y, s(u)), x, y, s(z), s(u), v) | quotIter#(x, s(y), z, u, v) → if#(le(x, z), x, s(y), z, u, v) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(x, 0) | → | quotZeroErro |
| quot(x, s(y)) | → | quotIter(x, s(y), 0, 0, 0) | | quotIter(x, s(y), z, u, v) | → | if(le(x, z), x, s(y), z, u, v) |
| if(true, x, y, z, u, v) | → | v | | if(false, x, y, z, u, v) | → | if2(le(y, s(u)), x, y, s(z), s(u), v) |
| if2(false, x, y, z, u, v) | → | quotIter(x, y, z, u, v) | | if2(true, x, y, z, u, v) | → | quotIter(x, y, z, 0, s(v)) |
Original Signature
Termination of terms over the following signature is verified: 0, le, s, quotZeroErro, quotIter, if, true, false, if2, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |