The TRS could be proven non-terminating. The proof took 307 ms.

The following reduction sequence is a witness for non-termination:

f#(g(___X), ___Y) →* f#(g(___X), ___Y)

The following DP Processors were used


Problem 1 was processed with processor SubtermCriterion (0ms).
 | – Problem 2 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 3 was processed with processor BackwardInstantiation (1ms).
 |    |    | – Problem 4 was processed with processor BackwardInstantiation (2ms).
 |    |    |    | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (1ms)].

Problem 1: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(g(X), Y)f#(X, f(g(X), Y))f#(g(X), Y)f#(g(X), Y)

Rewrite Rules

f(g(X), Y)f(X, f(g(X), Y))

Original Signature

Termination of terms over the following signature is verified: f, g

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(g(X), Y)f#(X, f(g(X), Y))

Problem 2: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(g(X), Y)f#(g(X), Y)

Rewrite Rules

f(g(X), Y)f(X, f(g(X), Y))

Original Signature

Termination of terms over the following signature is verified: f, g

Strategy


Instantiation

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

Problem 3: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(g(_X), _Y)f#(g(_X), _Y)

Rewrite Rules

f(g(X), Y)f(X, f(g(X), Y))

Original Signature

Termination of terms over the following signature is verified: f, g

Strategy


Instantiation

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

Problem 4: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(g(__X), __Y)f#(g(__X), __Y)

Rewrite Rules

f(g(X), Y)f(X, f(g(X), Y))

Original Signature

Termination of terms over the following signature is verified: f, g

Strategy


Instantiation

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