NO
The TRS could be proven non-terminating. The proof took 212 ms.
The following reduction sequence is a witness for non-termination:
f#(___x) →* f#(___x)
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (8ms).
| – Problem 2 was processed with processor BackwardInstantiation (1ms).
| | – Problem 3 was processed with processor BackwardInstantiation (1ms).
| | | – Problem 4 was processed with processor BackwardInstantiation (1ms).
| | | | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (3ms)].
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(x) | → | g#(f(x)) | | f#(x) | → | f#(x) |
Rewrite Rules
| f(x) | → | g(f(x)) | | g(f(x)) | → | x |
| g(x) | → | a |
Original Signature
Termination of terms over the following signature is verified: f, g, a
Strategy
The following SCCs where found
Problem 2: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(x) | → | g(f(x)) | | g(f(x)) | → | x |
| g(x) | → | a |
Original Signature
Termination of terms over the following signature is verified: f, g, a
Strategy
Instantiation
For all potential predecessors l → r of the rule f
#(
x) → f
#(
x) on dependency pair chains it holds that:
- f#(x) matches r,
- all variables of f#(x) are embedded in constructor contexts, i.e., each subterm of f#(x), containing a variable is rooted by a constructor symbol.
Thus, f
#(
x) → f
#(
x) is replaced by instances determined through the above matching. These instances are:
Problem 3: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(x) | → | g(f(x)) | | g(f(x)) | → | x |
| g(x) | → | a |
Original Signature
Termination of terms over the following signature is verified: f, g, a
Strategy
Instantiation
For all potential predecessors l → r of the rule f
#(
_x) → f
#(
_x) on dependency pair chains it holds that:
- f#(_x) matches r,
- all variables of f#(_x) are embedded in constructor contexts, i.e., each subterm of f#(_x), containing a variable is rooted by a constructor symbol.
Thus, f
#(
_x) → f
#(
_x) is replaced by instances determined through the above matching. These instances are:
Problem 4: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(x) | → | g(f(x)) | | g(f(x)) | → | x |
| g(x) | → | a |
Original Signature
Termination of terms over the following signature is verified: f, g, a
Strategy
Instantiation
For all potential predecessors l → r of the rule f
#(
__x) → f
#(
__x) on dependency pair chains it holds that:
- f#(__x) matches r,
- all variables of f#(__x) are embedded in constructor contexts, i.e., each subterm of f#(__x), containing a variable is rooted by a constructor symbol.
Thus, f
#(
__x) → f
#(
__x) is replaced by instances determined through the above matching. These instances are: