MAYBE
The TRS could not be proven terminating. The proof attempt took 430 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 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (52ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (34ms), DependencyGraph (2ms), ReductionPairSAT (25ms), DependencyGraph (1ms), SizeChangePrinciple (8ms)].
| – 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 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 5
Dependency Pairs
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, sel, first, cons, nil
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) | | quote#(s(X)) | → | quote#(X) |
| unquote#(s1(X)) | → | unquote#(X) | | first1#(s(X), cons(Y, Z)) | → | quote#(Y) |
| quote#(sel(X, Z)) | → | sel1#(X, Z) | | unquote1#(cons1(X, Z)) | → | unquote#(X) |
| unquote1#(cons1(X, Z)) | → | fcons#(unquote(X), unquote1(Z)) | | quote1#(cons(X, Z)) | → | quote#(X) |
| quote1#(first(X, Z)) | → | first1#(X, Z) | | first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
| unquote1#(cons1(X, Z)) | → | unquote1#(Z) | | sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) |
| first1#(s(X), cons(Y, Z)) | → | first1#(X, Z) | | from#(X) | → | from#(s(X)) |
| quote1#(cons(X, Z)) | → | quote1#(Z) | | sel1#(0, cons(X, Z)) | → | quote#(X) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
The following SCCs where found
| first#(s(X), cons(Y, Z)) → first#(X, Z) |
| unquote1#(cons1(X, Z)) → unquote1#(Z) |
| sel#(s(X), cons(Y, Z)) → sel#(X, Z) |
| first1#(s(X), cons(Y, Z)) → first1#(X, Z) |
| sel1#(s(X), cons(Y, Z)) → sel1#(X, Z) | quote#(s(X)) → quote#(X) |
| quote#(sel(X, Z)) → sel1#(X, Z) | sel1#(0, cons(X, Z)) → quote#(X) |
| unquote#(s1(X)) → unquote#(X) |
| quote1#(cons(X, Z)) → quote1#(Z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| unquote#(s1(X)) | → | unquote#(X) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| unquote#(s1(X)) | → | unquote#(X) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| first1#(s(X), cons(Y, Z)) | → | first1#(X, Z) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| first1#(s(X), cons(Y, Z)) | → | first1#(X, Z) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| unquote1#(cons1(X, Z)) | → | unquote1#(Z) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| unquote1#(cons1(X, Z)) | → | unquote1#(Z) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| quote1#(cons(X, Z)) | → | quote1#(Z) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| quote1#(cons(X, Z)) | → | quote1#(Z) |
Problem 7: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| first#(s(X), cons(Y, Z)) | → | first#(X, Z) |
Problem 8: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) |
Problem 9: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) | | quote#(s(X)) | → | quote#(X) |
| quote#(sel(X, Z)) | → | sel1#(X, Z) | | sel1#(0, cons(X, Z)) | → | quote#(X) |
Rewrite Rules
| sel(s(X), cons(Y, Z)) | → | sel(X, Z) | | sel(0, cons(X, Z)) | → | X |
| first(0, Z) | → | nil | | first(s(X), cons(Y, Z)) | → | cons(Y, first(X, Z)) |
| from(X) | → | cons(X, from(s(X))) | | sel1(s(X), cons(Y, Z)) | → | sel1(X, Z) |
| sel1(0, cons(X, Z)) | → | quote(X) | | first1(0, Z) | → | nil1 |
| first1(s(X), cons(Y, Z)) | → | cons1(quote(Y), first1(X, Z)) | | quote(0) | → | 01 |
| quote1(cons(X, Z)) | → | cons1(quote(X), quote1(Z)) | | quote1(nil) | → | nil1 |
| quote(s(X)) | → | s1(quote(X)) | | quote(sel(X, Z)) | → | sel1(X, Z) |
| quote1(first(X, Z)) | → | first1(X, Z) | | unquote(01) | → | 0 |
| unquote(s1(X)) | → | s(unquote(X)) | | unquote1(nil1) | → | nil |
| unquote1(cons1(X, Z)) | → | fcons(unquote(X), unquote1(Z)) | | fcons(X, Z) | → | cons(X, Z) |
Original Signature
Termination of terms over the following signature is verified: s1, cons1, fcons, from, 01, first1, quote1, unquote, 0, s, sel1, unquote1, quote, nil1, first, sel, nil, cons
Strategy
Projection
The following projection was used:
- π (quote#): 1
- π (sel1#): 2
Thus, the following dependency pairs are removed:
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) | | quote#(s(X)) | → | quote#(X) |
| quote#(sel(X, Z)) | → | sel1#(X, Z) | | sel1#(0, cons(X, Z)) | → | quote#(X) |