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 (310ms).
| – Problem 2 remains open; application of the following processors failed [PolynomialLinearRange4 (283ms), DependencyGraph (1ms), ReductionPairSAT (516ms), DependencyGraph (1ms), SizeChangePrinciple (53335ms)].
| – Problem 3 remains open; application of the following processors failed [PolynomialLinearRange4 (177ms), DependencyGraph (1ms), ReductionPairSAT (514ms), DependencyGraph (1ms), SizeChangePrinciple (timeout)].
| – Problem 4 remains open; application of the following processors failed [PolynomialLinearRange4 (279ms), DependencyGraph (51ms), ReductionPairSAT (570ms), DependencyGraph (55ms)].
| – Problem 5 remains open; application of the following processors failed [PolynomialLinearRange4 (181ms), DependencyGraph (1ms), ReductionPairSAT (440ms), DependencyGraph (1ms)].
The following open problems remain:
Open Dependency Pair Problem 2
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
Open Dependency Pair Problem 3
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
Open Dependency Pair Problem 4
Dependency Pairs
| T(sel(x_1, x_2)) | → | T(x_2) | | T(indx(x_1, x_2)) | → | T(x_1) |
| T(s(x_1)) | → | T(x_1) | | sel#(s(X), cons(Y, Z)) | → | T(Z) |
| sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) | | sel#(s(X), cons(Y, Z)) | → | T(X) |
| T(sel(x_1, x_2)) | → | T(x_1) | | T(dbl(x_1)) | → | T(x_1) |
| T(sel(X, Z)) | → | sel#(X, Z) | | T(dbls(x_1)) | → | T(x_1) |
| T(indx(x_1, x_2)) | → | T(x_2) | | sel#(0, cons(X, Y)) | → | T(X) |
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
Open Dependency Pair Problem 5
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, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| T(indx(x_1, x_2)) | → | T(x_1) | | quote#(s(X)) | → | quote#(X) |
| T(indx(Y, Z)) | → | indx#(Y, Z) | | T(from(s(X))) | → | from#(s(X)) |
| sel1#(s(X), cons(Y, Z)) | → | T(X) | | dbl1#(s(X)) | → | T(X) |
| sel1#(s(X), cons(Y, Z)) | → | T(Z) | | quote#(sel(X, Y)) | → | sel1#(X, Y) |
| dbl1#(s(X)) | → | dbl1#(X) | | quote#(dbl(X)) | → | dbl1#(X) |
| sel#(s(X), cons(Y, Z)) | → | sel#(X, Z) | | sel1#(0, cons(X, Y)) | → | T(X) |
| quote#(s(X)) | → | T(X) | | T(sel(x_1, x_2)) | → | T(x_1) |
| sel1#(s(X), cons(Y, Z)) | → | sel1#(X, Z) | | T(dbl(x_1)) | → | T(x_1) |
| sel#(0, cons(X, Y)) | → | T(X) | | T(sel(x_1, x_2)) | → | T(x_2) |
| T(s(x_1)) | → | T(x_1) | | sel#(s(X), cons(Y, Z)) | → | T(Z) |
| T(dbl(X)) | → | dbl#(X) | | sel#(s(X), cons(Y, Z)) | → | T(X) |
| T(dbls(Y)) | → | dbls#(Y) | | T(dbls(x_1)) | → | T(x_1) |
| T(sel(X, Z)) | → | sel#(X, Z) | | T(indx(x_1, x_2)) | → | T(x_2) |
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
Context-sensitive strategy:
μ(from#) = μ(from) = μ(01) = μ(T) = μ(0) = μ(s) = μ(nil) = μ(cons) = ∅
μ(quote#) = μ(s1) = μ(dbl1) = μ(dbl) = μ(dbls) = μ(dbls#) = μ(dbl#) = μ(indx) = μ(quote) = μ(dbl1#) = μ(indx#) = {1}
μ(sel#) = μ(sel1#) = μ(sel1) = μ(sel) = {1, 2}
The following SCCs where found
| sel#(s(X), cons(Y, Z)) → T(X) | sel#(s(X), cons(Y, Z)) → sel#(X, Z) |
| T(sel(x_1, x_2)) → T(x_2) | T(indx(x_1, x_2)) → T(x_1) |
| T(s(x_1)) → T(x_1) | T(sel(x_1, x_2)) → T(x_1) |
| sel#(s(X), cons(Y, Z)) → T(Z) | T(dbl(x_1)) → T(x_1) |
| T(dbls(x_1)) → T(x_1) | T(sel(X, Z)) → sel#(X, Z) |
| sel#(0, cons(X, Y)) → T(X) | T(indx(x_1, x_2)) → T(x_2) |
| sel1#(s(X), cons(Y, Z)) → sel1#(X, Z) |