NO
The TRS could be proven non-terminating. The proof took 1363 ms.
The following reduction sequence is a witness for non-termination:
f#(g(0, 0)) →* f#(g(0, 0))
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| – Problem 2 was processed with processor BackwardsNarrowing (1ms).
| | – Problem 3 was processed with processor BackwardsNarrowing (1ms).
| | | – Problem 4 remains open; application of the following processors failed [BackwardsNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms), BackwardsNarrowing (0ms), BackwardInstantiation (0ms), ForwardInstantiation (0ms), Propagation (2ms)].
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x)) | → | g#(x, x) | | g#(0, 1) | → | 0# |
| f#(s(x)) | → | f#(g(x, x)) |
Rewrite Rules
| f(s(x)) | → | f(g(x, x)) | | g(0, 1) | → | s(0) |
| 0 | → | 1 |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s
Strategy
The following SCCs where found
Problem 2: BackwardsNarrowing
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(s(x)) | → | f(g(x, x)) | | g(0, 1) | → | s(0) |
| 0 | → | 1 |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s
Strategy
The left-hand side of the rule f
#(s(
x)) → f
#(g(
x,
x)) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
| Relevant Terms | Irrelevant Terms |
|---|
| f#(g(0, 1)) | |
Thus, the rule f
#(s(
x)) → f
#(g(
x,
x)) is replaced by the following rules:
| f#(g(0, 1)) → f#(g(0, 0)) |
Problem 3: BackwardsNarrowing
Dependency Pair Problem
Dependency Pairs
| f#(g(0, 1)) | → | f#(g(0, 0)) |
Rewrite Rules
| f(s(x)) | → | f(g(x, x)) | | g(0, 1) | → | s(0) |
| 0 | → | 1 |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0, s
Strategy
The left-hand side of the rule f
#(g(0, 1)) → f
#(g(0, 0)) is backward narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
| Relevant Terms | Irrelevant Terms |
|---|
| f#(g(0, 0)) | |
Thus, the rule f
#(g(0, 1)) → f
#(g(0, 0)) is replaced by the following rules:
| f#(g(0, 0)) → f#(g(0, 0)) |