(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(a)) → a
f(f(x)) → b
g(x) → f(g(x))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a'(g(f(x))) → a'(x)
f(f(x)) → b'(x)
g(x) → g(f(x))
Q is empty.
(3) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a'(g(f(x))) → a'(x)
f(f(x)) → b'(x)
g(x) → g(f(x))
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a) = 0
POL(b) = 0
POL(f(x1)) = x1
POL(g(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(g(a)) → a
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x)) → b
g(x) → f(g(x))
Q is empty.
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(x) → F(g(x))
G(x) → G(x)
The TRS R consists of the following rules:
f(f(x)) → b
g(x) → f(g(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(x) → G(x)
The TRS R consists of the following rules:
f(f(x)) → b
g(x) → f(g(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(x) → G(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(x) → G(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
G(
x) evaluates to t =
G(
x)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from G(x) to G(x).
(16) FALSE