MAYBE
The TRS could not be proven terminating. The proof attempt took 407 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 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor SubtermCriterion (0ms).
| – Problem 6 was processed with processor SubtermCriterion (0ms).
| – Problem 7 was processed with processor SubtermCriterion (0ms).
| – Problem 8 was processed with processor SubtermCriterion (0ms).
| – Problem 9 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (53ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (35ms), DependencyGraph (1ms), ReductionPairSAT (25ms), DependencyGraph (1ms), SizeChangePrinciple (7ms)].
The following open problems remain:
Open Dependency Pair Problem 9
Dependency Pairs
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) | | quote#(s(X)) | → | quote#(X) |
| dbl1#(s(X)) | → | dbl1#(X) | | sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) |
| dbl#(s(X)) | → | dbl#(X) | | indx#(cons(X, Y), Z) | → | sel#(X, Z) |
| from#(X) | → | from#(s(X)) | | dbls#(cons(X, Y)) | → | dbls#(Y) |
| dbls#(cons(X, Y)) | → | dbl#(X) | | indx#(cons(X, Y), Z) | → | indx#(Y, Z) |
| quote#(dbl(X)) | → | dbl1#(X) | | quote#(sel(X, Y)) | → | sel1#(X, Y) |
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
The following SCCs where found
| sel1#(s(X), cons(Y, Z)) → sel1#(X, Z) |
| sel#(s(X), cons(Y, Z)) → sel#(X, Z) |
| dbls#(cons(X, Y)) → dbls#(Y) |
| indx#(cons(X, Y), Z) → indx#(Y, Z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) |
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| dbls#(cons(X, Y)) | → | dbls#(Y) |
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| dbls#(cons(X, Y)) | → | dbls#(Y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) |
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| indx#(cons(X, Y), Z) | → | indx#(Y, Z) |
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| indx#(cons(X, Y), Z) | → | indx#(Y, Z) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 8: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| dbl(0) | → | 0 | | dbl(s(X)) | → | s(s(dbl(X))) |
| dbls(nil) | → | nil | | dbls(cons(X, Y)) | → | cons(dbl(X), dbls(Y)) |
| sel(0, cons(X, Y)) | → | X | | sel(s(X), cons(Y, Z)) | → | sel(X, Z) |
| indx(nil, X) | → | nil | | indx(cons(X, Y), Z) | → | cons(sel(X, Z), indx(Y, Z)) |
| from(X) | → | cons(X, from(s(X))) | | dbl1(0) | → | 01 |
| dbl1(s(X)) | → | s1(s1(dbl1(X))) | | sel1(0, cons(X, Y)) | → | X |
| sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) | | quote(0) | → | 01 |
| quote(s(X)) | → | s1(quote(X)) | | quote(dbl(X)) | → | dbl1(X) |
| quote(sel(X, Y)) | → | sel1(X, Y) |
Original Signature
Termination of terms over the following signature is verified: s1, dbl1, dbl, from, 01, dbls, 0, s, indx, sel1, quote, sel, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: