NO

The TRS could be proven non-terminating. The proof took 261 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 (1ms).
 |    | – Problem 3 was processed with processor BackwardInstantiation (1ms).
 |    |    | – Problem 4 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (2ms)].

Problem 1: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(x)f#(x)

Rewrite Rules

f(x)f(x)

Original Signature

Termination of terms over the following signature is verified: f

Strategy


Instantiation

For all potential predecessors l → r of the rule f#(x) → f#(x) on dependency pair chains it holds that: Thus, f#(x) → f#(x) is replaced by instances determined through the above matching. These instances are:
f#(_x) → f#(_x)

Problem 2: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(_x)f#(_x)

Rewrite Rules

f(x)f(x)

Original Signature

Termination of terms over the following signature is verified: f

Strategy


Instantiation

For all potential predecessors l → r of the rule f#(_x) → f#(_x) on dependency pair chains it holds that: Thus, f#(_x) → f#(_x) is replaced by instances determined through the above matching. These instances are:
f#(__x) → f#(__x)

Problem 3: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(__x)f#(__x)

Rewrite Rules

f(x)f(x)

Original Signature

Termination of terms over the following signature is verified: f

Strategy


Instantiation

For all potential predecessors l → r of the rule f#(__x) → f#(__x) on dependency pair chains it holds that: Thus, f#(__x) → f#(__x) is replaced by instances determined through the above matching. These instances are:
f#(___x) → f#(___x)