TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60001 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (45ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (8ms), PolynomialLinearRange4iUR (440ms), DependencyGraph (6ms), PolynomialLinearRange8NegiUR (8206ms), DependencyGraph (4ms), ReductionPairSAT (timeout)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| if2#(true, x, y, z, u) | → | modIter#(x, y, z, 0) | | modIter#(x, s(y), z, u) | → | if#(le(x, z), x, s(y), z, u) |
| if#(false, x, y, z, u) | → | if2#(le(y, s(u)), x, y, s(z), s(u)) | | if2#(false, x, y, z, u) | → | modIter#(x, y, z, u) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | mod(x, 0) | → | modZeroErro |
| mod(x, s(y)) | → | modIter(x, s(y), 0, 0) | | modIter(x, s(y), z, u) | → | if(le(x, z), x, s(y), z, u) |
| if(true, x, y, z, u) | → | u | | if(false, x, y, z, u) | → | if2(le(y, s(u)), x, y, s(z), s(u)) |
| if2(false, x, y, z, u) | → | modIter(x, y, z, u) | | if2(true, x, y, z, u) | → | modIter(x, y, z, 0) |
Original Signature
Termination of terms over the following signature is verified: modIter, 0, s, le, if, mod, false, true, if2, modZeroErro
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| if2#(true, x, y, z, u) | → | modIter#(x, y, z, 0) | | le#(s(x), s(y)) | → | le#(x, y) |
| modIter#(x, s(y), z, u) | → | le#(x, z) | | modIter#(x, s(y), z, u) | → | if#(le(x, z), x, s(y), z, u) |
| if#(false, x, y, z, u) | → | if2#(le(y, s(u)), x, y, s(z), s(u)) | | if#(false, x, y, z, u) | → | le#(y, s(u)) |
| mod#(x, s(y)) | → | modIter#(x, s(y), 0, 0) | | if2#(false, x, y, z, u) | → | modIter#(x, y, z, u) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | mod(x, 0) | → | modZeroErro |
| mod(x, s(y)) | → | modIter(x, s(y), 0, 0) | | modIter(x, s(y), z, u) | → | if(le(x, z), x, s(y), z, u) |
| if(true, x, y, z, u) | → | u | | if(false, x, y, z, u) | → | if2(le(y, s(u)), x, y, s(z), s(u)) |
| if2(false, x, y, z, u) | → | modIter(x, y, z, u) | | if2(true, x, y, z, u) | → | modIter(x, y, z, 0) |
Original Signature
Termination of terms over the following signature is verified: modIter, 0, le, s, if, mod, true, false, if2, modZeroErro
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| if2#(true, x, y, z, u) → modIter#(x, y, z, 0) | modIter#(x, s(y), z, u) → if#(le(x, z), x, s(y), z, u) |
| if#(false, x, y, z, u) → if2#(le(y, s(u)), x, y, s(z), s(u)) | if2#(false, x, y, z, u) → modIter#(x, y, z, u) |
Problem 3: 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) | | mod(x, 0) | → | modZeroErro |
| mod(x, s(y)) | → | modIter(x, s(y), 0, 0) | | modIter(x, s(y), z, u) | → | if(le(x, z), x, s(y), z, u) |
| if(true, x, y, z, u) | → | u | | if(false, x, y, z, u) | → | if2(le(y, s(u)), x, y, s(z), s(u)) |
| if2(false, x, y, z, u) | → | modIter(x, y, z, u) | | if2(true, x, y, z, u) | → | modIter(x, y, z, 0) |
Original Signature
Termination of terms over the following signature is verified: modIter, 0, le, s, if, mod, true, false, if2, modZeroErro
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |