TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60066 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (43ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (724ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (5956ms), DependencyGraph (2ms), ReductionPairSAT (53217ms), DependencyGraph (2ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| f#(true, x, y, z) | → | f#(and(gt(x, y), gt(x, z)), x, y, s(z)) | | f#(true, x, y, z) | → | f#(and(gt(x, y), gt(x, z)), x, s(y), z) |
Rewrite Rules
| f(true, x, y, z) | → | f(and(gt(x, y), gt(x, z)), x, s(y), z) | | f(true, x, y, z) | → | f(and(gt(x, y), gt(x, z)), x, y, s(z)) |
| gt(0, v) | → | false | | gt(s(u), 0) | → | true |
| gt(s(u), s(v)) | → | gt(u, v) | | and(x, true) | → | x |
| and(x, false) | → | false |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, false, true, gt, and
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(true, x, y, z) | → | f#(and(gt(x, y), gt(x, z)), x, s(y), z) | | f#(true, x, y, z) | → | f#(and(gt(x, y), gt(x, z)), x, y, s(z)) |
| gt#(s(u), s(v)) | → | gt#(u, v) | | f#(true, x, y, z) | → | gt#(x, y) |
| f#(true, x, y, z) | → | gt#(x, z) | | f#(true, x, y, z) | → | and#(gt(x, y), gt(x, z)) |
Rewrite Rules
| f(true, x, y, z) | → | f(and(gt(x, y), gt(x, z)), x, s(y), z) | | f(true, x, y, z) | → | f(and(gt(x, y), gt(x, z)), x, y, s(z)) |
| gt(0, v) | → | false | | gt(s(u), 0) | → | true |
| gt(s(u), s(v)) | → | gt(u, v) | | and(x, true) | → | x |
| and(x, false) | → | false |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, true, false, gt, and
Strategy
The following SCCs where found
| f#(true, x, y, z) → f#(and(gt(x, y), gt(x, z)), x, s(y), z) | f#(true, x, y, z) → f#(and(gt(x, y), gt(x, z)), x, y, s(z)) |
| gt#(s(u), s(v)) → gt#(u, v) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gt#(s(u), s(v)) | → | gt#(u, v) |
Rewrite Rules
| f(true, x, y, z) | → | f(and(gt(x, y), gt(x, z)), x, s(y), z) | | f(true, x, y, z) | → | f(and(gt(x, y), gt(x, z)), x, y, s(z)) |
| gt(0, v) | → | false | | gt(s(u), 0) | → | true |
| gt(s(u), s(v)) | → | gt(u, v) | | and(x, true) | → | x |
| and(x, false) | → | false |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, true, false, gt, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gt#(s(u), s(v)) | → | gt#(u, v) |