NO
The TRS could be proven non-terminating. The proof took 239 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 BackwardInstantiation (1ms).
| – Problem 2 was processed with processor BackwardInstantiation (3ms).
| | – Problem 3 was processed with processor BackwardInstantiation (2ms).
| | | – Problem 4 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (1ms)].
Problem 1: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
Original Signature
Termination of terms over the following signature is verified: f, g, h
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 2: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
Original Signature
Termination of terms over the following signature is verified: f, g, h
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
Original Signature
Termination of terms over the following signature is verified: f, g, h
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: