TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60000 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (79ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (752ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (3183ms), DependencyGraph (0ms), ReductionPairSAT (1228ms), DependencyGraph (1ms), SizeChangePrinciple (20514ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
| – Problem 6 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 5
Dependency Pairs
| minsort#(cons(x, y)) | → | minsort#(del(min(x, y), cons(x, y))) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | eq(0, 0) | → | true |
| eq(0, s(y)) | → | false | | eq(s(x), 0) | → | false |
| eq(s(x), s(y)) | → | eq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | minsort(nil) | → | nil |
| minsort(cons(x, y)) | → | cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) | | min(x, nil) | → | x |
| min(x, cons(y, z)) | → | if(le(x, y), min(x, z), min(y, z)) | | del(x, nil) | → | nil |
| del(x, cons(y, z)) | → | if(eq(x, y), z, cons(y, del(x, z))) |
Original Signature
Termination of terms over the following signature is verified: min, minsort, 0, s, le, if, false, true, del, cons, nil, eq
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minsort#(cons(x, y)) | → | minsort#(del(min(x, y), cons(x, y))) | | del#(x, cons(y, z)) | → | if#(eq(x, y), z, cons(y, del(x, z))) |
| del#(x, cons(y, z)) | → | del#(x, z) | | le#(s(x), s(y)) | → | le#(x, y) |
| minsort#(cons(x, y)) | → | min#(x, y) | | min#(x, cons(y, z)) | → | if#(le(x, y), min(x, z), min(y, z)) |
| min#(x, cons(y, z)) | → | min#(y, z) | | del#(x, cons(y, z)) | → | eq#(x, y) |
| min#(x, cons(y, z)) | → | min#(x, z) | | eq#(s(x), s(y)) | → | eq#(x, y) |
| min#(x, cons(y, z)) | → | le#(x, y) | | minsort#(cons(x, y)) | → | del#(min(x, y), cons(x, y)) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | eq(0, 0) | → | true |
| eq(0, s(y)) | → | false | | eq(s(x), 0) | → | false |
| eq(s(x), s(y)) | → | eq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | minsort(nil) | → | nil |
| minsort(cons(x, y)) | → | cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) | | min(x, nil) | → | x |
| min(x, cons(y, z)) | → | if(le(x, y), min(x, z), min(y, z)) | | del(x, nil) | → | nil |
| del(x, cons(y, z)) | → | if(eq(x, y), z, cons(y, del(x, z))) |
Original Signature
Termination of terms over the following signature is verified: minsort, min, 0, le, s, if, true, false, del, eq, nil, cons
Strategy
The following SCCs where found
| minsort#(cons(x, y)) → minsort#(del(min(x, y), cons(x, y))) |
| del#(x, cons(y, z)) → del#(x, z) |
| le#(s(x), s(y)) → le#(x, y) |
| min#(x, cons(y, z)) → min#(y, z) | min#(x, cons(y, z)) → min#(x, z) |
| eq#(s(x), s(y)) → eq#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| eq#(s(x), s(y)) | → | eq#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | eq(0, 0) | → | true |
| eq(0, s(y)) | → | false | | eq(s(x), 0) | → | false |
| eq(s(x), s(y)) | → | eq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | minsort(nil) | → | nil |
| minsort(cons(x, y)) | → | cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) | | min(x, nil) | → | x |
| min(x, cons(y, z)) | → | if(le(x, y), min(x, z), min(y, z)) | | del(x, nil) | → | nil |
| del(x, cons(y, z)) | → | if(eq(x, y), z, cons(y, del(x, z))) |
Original Signature
Termination of terms over the following signature is verified: minsort, min, 0, le, s, if, true, false, del, eq, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| eq#(s(x), s(y)) | → | eq#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| del#(x, cons(y, z)) | → | del#(x, z) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | eq(0, 0) | → | true |
| eq(0, s(y)) | → | false | | eq(s(x), 0) | → | false |
| eq(s(x), s(y)) | → | eq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | minsort(nil) | → | nil |
| minsort(cons(x, y)) | → | cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) | | min(x, nil) | → | x |
| min(x, cons(y, z)) | → | if(le(x, y), min(x, z), min(y, z)) | | del(x, nil) | → | nil |
| del(x, cons(y, z)) | → | if(eq(x, y), z, cons(y, del(x, z))) |
Original Signature
Termination of terms over the following signature is verified: minsort, min, 0, le, s, if, true, false, del, eq, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| del#(x, cons(y, z)) | → | del#(x, z) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| min#(x, cons(y, z)) | → | min#(y, z) | | min#(x, cons(y, z)) | → | min#(x, z) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | eq(0, 0) | → | true |
| eq(0, s(y)) | → | false | | eq(s(x), 0) | → | false |
| eq(s(x), s(y)) | → | eq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | minsort(nil) | → | nil |
| minsort(cons(x, y)) | → | cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) | | min(x, nil) | → | x |
| min(x, cons(y, z)) | → | if(le(x, y), min(x, z), min(y, z)) | | del(x, nil) | → | nil |
| del(x, cons(y, z)) | → | if(eq(x, y), z, cons(y, del(x, z))) |
Original Signature
Termination of terms over the following signature is verified: minsort, min, 0, le, s, if, true, false, del, eq, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| min#(x, cons(y, z)) | → | min#(y, z) | | min#(x, cons(y, z)) | → | min#(x, z) |
Problem 6: 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) | | eq(0, 0) | → | true |
| eq(0, s(y)) | → | false | | eq(s(x), 0) | → | false |
| eq(s(x), s(y)) | → | eq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | minsort(nil) | → | nil |
| minsort(cons(x, y)) | → | cons(min(x, y), minsort(del(min(x, y), cons(x, y)))) | | min(x, nil) | → | x |
| min(x, cons(y, z)) | → | if(le(x, y), min(x, z), min(y, z)) | | del(x, nil) | → | nil |
| del(x, cons(y, z)) | → | if(eq(x, y), z, cons(y, del(x, z))) |
Original Signature
Termination of terms over the following signature is verified: minsort, min, 0, le, s, if, true, false, del, eq, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |