TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60062 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (88ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (463ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (3421ms), DependencyGraph (3ms), ReductionPairSAT (timeout)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| if#(first, x, y, z) | → | if#(le(s(x), y, s(z)), s(x), y, s(z)) | | if#(second, x, y, z) | → | if#(le(s(x), s(y), z), s(x), s(y), z) |
Rewrite Rules
| le(0, y, z) | → | greater(y, z) | | le(s(x), 0, z) | → | false |
| le(s(x), s(y), 0) | → | false | | le(s(x), s(y), s(z)) | → | le(x, y, z) |
| greater(x, 0) | → | first | | greater(0, s(y)) | → | second |
| greater(s(x), s(y)) | → | greater(x, y) | | double(0) | → | 0 |
| double(s(x)) | → | s(s(double(x))) | | triple(x) | → | if(le(x, x, double(x)), x, 0, 0) |
| if(false, x, y, z) | → | true | | if(first, x, y, z) | → | if(le(s(x), y, s(z)), s(x), y, s(z)) |
| if(second, x, y, z) | → | if(le(s(x), s(y), z), s(x), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: 0, greater, second, s, le, if, triple, true, false, double, first
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| triple#(x) | → | le#(x, x, double(x)) | | le#(s(x), s(y), s(z)) | → | le#(x, y, z) |
| triple#(x) | → | double#(x) | | if#(first, x, y, z) | → | if#(le(s(x), y, s(z)), s(x), y, s(z)) |
| if#(second, x, y, z) | → | le#(s(x), s(y), z) | | double#(s(x)) | → | double#(x) |
| le#(0, y, z) | → | greater#(y, z) | | greater#(s(x), s(y)) | → | greater#(x, y) |
| triple#(x) | → | if#(le(x, x, double(x)), x, 0, 0) | | if#(first, x, y, z) | → | le#(s(x), y, s(z)) |
| if#(second, x, y, z) | → | if#(le(s(x), s(y), z), s(x), s(y), z) |
Rewrite Rules
| le(0, y, z) | → | greater(y, z) | | le(s(x), 0, z) | → | false |
| le(s(x), s(y), 0) | → | false | | le(s(x), s(y), s(z)) | → | le(x, y, z) |
| greater(x, 0) | → | first | | greater(0, s(y)) | → | second |
| greater(s(x), s(y)) | → | greater(x, y) | | double(0) | → | 0 |
| double(s(x)) | → | s(s(double(x))) | | triple(x) | → | if(le(x, x, double(x)), x, 0, 0) |
| if(false, x, y, z) | → | true | | if(first, x, y, z) | → | if(le(s(x), y, s(z)), s(x), y, s(z)) |
| if(second, x, y, z) | → | if(le(s(x), s(y), z), s(x), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first
Strategy
The following SCCs where found
| le#(s(x), s(y), s(z)) → le#(x, y, z) |
| if#(first, x, y, z) → if#(le(s(x), y, s(z)), s(x), y, s(z)) | if#(second, x, y, z) → if#(le(s(x), s(y), z), s(x), s(y), z) |
| double#(s(x)) → double#(x) |
| greater#(s(x), s(y)) → greater#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| greater#(s(x), s(y)) | → | greater#(x, y) |
Rewrite Rules
| le(0, y, z) | → | greater(y, z) | | le(s(x), 0, z) | → | false |
| le(s(x), s(y), 0) | → | false | | le(s(x), s(y), s(z)) | → | le(x, y, z) |
| greater(x, 0) | → | first | | greater(0, s(y)) | → | second |
| greater(s(x), s(y)) | → | greater(x, y) | | double(0) | → | 0 |
| double(s(x)) | → | s(s(double(x))) | | triple(x) | → | if(le(x, x, double(x)), x, 0, 0) |
| if(false, x, y, z) | → | true | | if(first, x, y, z) | → | if(le(s(x), y, s(z)), s(x), y, s(z)) |
| if(second, x, y, z) | → | if(le(s(x), s(y), z), s(x), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| greater#(s(x), s(y)) | → | greater#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y), s(z)) | → | le#(x, y, z) |
Rewrite Rules
| le(0, y, z) | → | greater(y, z) | | le(s(x), 0, z) | → | false |
| le(s(x), s(y), 0) | → | false | | le(s(x), s(y), s(z)) | → | le(x, y, z) |
| greater(x, 0) | → | first | | greater(0, s(y)) | → | second |
| greater(s(x), s(y)) | → | greater(x, y) | | double(0) | → | 0 |
| double(s(x)) | → | s(s(double(x))) | | triple(x) | → | if(le(x, x, double(x)), x, 0, 0) |
| if(false, x, y, z) | → | true | | if(first, x, y, z) | → | if(le(s(x), y, s(z)), s(x), y, s(z)) |
| if(second, x, y, z) | → | if(le(s(x), s(y), z), s(x), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y), s(z)) | → | le#(x, y, z) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| double#(s(x)) | → | double#(x) |
Rewrite Rules
| le(0, y, z) | → | greater(y, z) | | le(s(x), 0, z) | → | false |
| le(s(x), s(y), 0) | → | false | | le(s(x), s(y), s(z)) | → | le(x, y, z) |
| greater(x, 0) | → | first | | greater(0, s(y)) | → | second |
| greater(s(x), s(y)) | → | greater(x, y) | | double(0) | → | 0 |
| double(s(x)) | → | s(s(double(x))) | | triple(x) | → | if(le(x, x, double(x)), x, 0, 0) |
| if(false, x, y, z) | → | true | | if(first, x, y, z) | → | if(le(s(x), y, s(z)), s(x), y, s(z)) |
| if(second, x, y, z) | → | if(le(s(x), s(y), z), s(x), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| double#(s(x)) | → | double#(x) |