YES
The TRS could be proven terminating. The proof took 36 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (21ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) | | fib#(N) | → | sel#(N, fib1(s(0), s(0))) |
| activate#(n__fib1(X1, X2)) | → | fib1#(X1, X2) | | fib#(N) | → | fib1#(s(0), s(0)) |
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) | | sel#(s(N), cons(X, XS)) | → | activate#(XS) |
| fib1#(X, Y) | → | add#(X, Y) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, n__fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| sel(0, cons(X, XS)) | → | X | | sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) |
| fib1(X1, X2) | → | n__fib1(X1, X2) | | activate(n__fib1(X1, X2)) | → | fib1(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 0, s, fib1, add, sel, fib, n__fib1, cons
Strategy
The following SCCs where found
| add#(s(X), Y) → add#(X, Y) |
| sel#(s(N), cons(X, XS)) → sel#(N, activate(XS)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, n__fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| sel(0, cons(X, XS)) | → | X | | sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) |
| fib1(X1, X2) | → | n__fib1(X1, X2) | | activate(n__fib1(X1, X2)) | → | fib1(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 0, s, fib1, add, sel, fib, n__fib1, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| sel#(s(N), cons(X, XS)) | → | sel#(N, activate(XS)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| add#(s(X), Y) | → | add#(X, Y) |
Rewrite Rules
| fib(N) | → | sel(N, fib1(s(0), s(0))) | | fib1(X, Y) | → | cons(X, n__fib1(Y, add(X, Y))) |
| add(0, X) | → | X | | add(s(X), Y) | → | s(add(X, Y)) |
| sel(0, cons(X, XS)) | → | X | | sel(s(N), cons(X, XS)) | → | sel(N, activate(XS)) |
| fib1(X1, X2) | → | n__fib1(X1, X2) | | activate(n__fib1(X1, X2)) | → | fib1(X1, X2) |
| activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: activate, 0, s, fib1, add, sel, fib, n__fib1, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| add#(s(X), Y) | → | add#(X, Y) |