MAYBE
The TRS could not be proven terminating. The proof attempt took 332 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (0ms), PolynomialLinearRange4iUR (72ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (34ms), DependencyGraph (1ms), ReductionPairSAT (22ms), DependencyGraph (1ms), SizeChangePrinciple (5ms)].
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
Rewrite Rules
| eq(0, 0) | → | true | | eq(s(X), s(Y)) | → | eq(X, Y) |
| eq(X, Y) | → | false | | inf(X) | → | cons(X, inf(s(X))) |
| take(0, X) | → | nil | | take(s(X), cons(Y, L)) | → | cons(Y, take(X, L)) |
| length(nil) | → | 0 | | length(cons(X, L)) | → | s(length(L)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, take, length, inf, false, true, nil, cons, eq
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| inf#(X) | → | inf#(s(X)) | | eq#(s(X), s(Y)) | → | eq#(X, Y) |
| take#(s(X), cons(Y, L)) | → | take#(X, L) | | length#(cons(X, L)) | → | length#(L) |
Rewrite Rules
| eq(0, 0) | → | true | | eq(s(X), s(Y)) | → | eq(X, Y) |
| eq(X, Y) | → | false | | inf(X) | → | cons(X, inf(s(X))) |
| take(0, X) | → | nil | | take(s(X), cons(Y, L)) | → | cons(Y, take(X, L)) |
| length(nil) | → | 0 | | length(cons(X, L)) | → | s(length(L)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, take, inf, length, true, false, eq, cons, nil
Strategy
The following SCCs where found
| eq#(s(X), s(Y)) → eq#(X, Y) |
| take#(s(X), cons(Y, L)) → take#(X, L) |
| length#(cons(X, L)) → length#(L) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| take#(s(X), cons(Y, L)) | → | take#(X, L) |
Rewrite Rules
| eq(0, 0) | → | true | | eq(s(X), s(Y)) | → | eq(X, Y) |
| eq(X, Y) | → | false | | inf(X) | → | cons(X, inf(s(X))) |
| take(0, X) | → | nil | | take(s(X), cons(Y, L)) | → | cons(Y, take(X, L)) |
| length(nil) | → | 0 | | length(cons(X, L)) | → | s(length(L)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, take, inf, length, true, false, eq, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| take#(s(X), cons(Y, L)) | → | take#(X, L) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| eq#(s(X), s(Y)) | → | eq#(X, Y) |
Rewrite Rules
| eq(0, 0) | → | true | | eq(s(X), s(Y)) | → | eq(X, Y) |
| eq(X, Y) | → | false | | inf(X) | → | cons(X, inf(s(X))) |
| take(0, X) | → | nil | | take(s(X), cons(Y, L)) | → | cons(Y, take(X, L)) |
| length(nil) | → | 0 | | length(cons(X, L)) | → | s(length(L)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, take, inf, length, true, false, eq, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| eq#(s(X), s(Y)) | → | eq#(X, Y) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| length#(cons(X, L)) | → | length#(L) |
Rewrite Rules
| eq(0, 0) | → | true | | eq(s(X), s(Y)) | → | eq(X, Y) |
| eq(X, Y) | → | false | | inf(X) | → | cons(X, inf(s(X))) |
| take(0, X) | → | nil | | take(s(X), cons(Y, L)) | → | cons(Y, take(X, L)) |
| length(nil) | → | 0 | | length(cons(X, L)) | → | s(length(L)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, take, inf, length, true, false, eq, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| length#(cons(X, L)) | → | length#(L) |