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 (34ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (873ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (4328ms), DependencyGraph (5ms), ReductionPairSAT (timeout)].
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| function#(if, false, x, y) | → | function#(plus, function(third, x, y, y), function(p, x, x, y), s(y)) | | function#(plus, dummy, x, y) | → | function#(if, function(iszero, x, x, x), x, y) |
Rewrite Rules
| function(iszero, 0, dummy, dummy2) | → | true | | function(iszero, s(x), dummy, dummy2) | → | false |
| function(p, 0, dummy, dummy2) | → | 0 | | function(p, s(0), dummy, dummy2) | → | 0 |
| function(p, s(s(x)), dummy, dummy2) | → | s(function(p, s(x), x, x)) | | function(plus, dummy, x, y) | → | function(if, function(iszero, x, x, x), x, y) |
| function(if, true, x, y) | → | y | | function(if, false, x, y) | → | function(plus, function(third, x, y, y), function(p, x, x, y), s(y)) |
| function(third, x, y, z) | → | z |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, if, p, false, true, iszero, third, function
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| function#(p, s(s(x)), dummy, dummy2) | → | function#(p, s(x), x, x) | | function#(if, false, x, y) | → | function#(plus, function(third, x, y, y), function(p, x, x, y), s(y)) |
| function#(plus, dummy, x, y) | → | function#(iszero, x, x, x) | | function#(plus, dummy, x, y) | → | function#(if, function(iszero, x, x, x), x, y) |
| function#(if, false, x, y) | → | function#(third, x, y, y) | | function#(if, false, x, y) | → | function#(p, x, x, y) |
Rewrite Rules
| function(iszero, 0, dummy, dummy2) | → | true | | function(iszero, s(x), dummy, dummy2) | → | false |
| function(p, 0, dummy, dummy2) | → | 0 | | function(p, s(0), dummy, dummy2) | → | 0 |
| function(p, s(s(x)), dummy, dummy2) | → | s(function(p, s(x), x, x)) | | function(plus, dummy, x, y) | → | function(if, function(iszero, x, x, x), x, y) |
| function(if, true, x, y) | → | y | | function(if, false, x, y) | → | function(plus, function(third, x, y, y), function(p, x, x, y), s(y)) |
| function(third, x, y, z) | → | z |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, if, p, true, false, iszero, third, function
Strategy
The following SCCs where found
| function#(if, false, x, y) → function#(plus, function(third, x, y, y), function(p, x, x, y), s(y)) | function#(plus, dummy, x, y) → function#(if, function(iszero, x, x, x), x, y) |
| function#(p, s(s(x)), dummy, dummy2) → function#(p, s(x), x, x) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| function#(p, s(s(x)), dummy, dummy2) | → | function#(p, s(x), x, x) |
Rewrite Rules
| function(iszero, 0, dummy, dummy2) | → | true | | function(iszero, s(x), dummy, dummy2) | → | false |
| function(p, 0, dummy, dummy2) | → | 0 | | function(p, s(0), dummy, dummy2) | → | 0 |
| function(p, s(s(x)), dummy, dummy2) | → | s(function(p, s(x), x, x)) | | function(plus, dummy, x, y) | → | function(if, function(iszero, x, x, x), x, y) |
| function(if, true, x, y) | → | y | | function(if, false, x, y) | → | function(plus, function(third, x, y, y), function(p, x, x, y), s(y)) |
| function(third, x, y, z) | → | z |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, if, p, true, false, iszero, third, function
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| function#(p, s(s(x)), dummy, dummy2) | → | function#(p, s(x), x, x) |