(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → g(f(x))
g(f(x)) → x
g(x) → a
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:
f(x) → f(g(x))
f(g(x)) → x
g(x) → a'(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:
f(x) → f(g(x))
f(g(x)) → x
g(x) → a'(x)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a) = 0
POL(f(x1)) = 1 + x1
POL(g(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
g(f(x)) → x
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → g(f(x))
g(x) → a
Q is empty.
(7) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(8) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → g(f(x))
g(x) → a
The set Q consists of the following terms:
f(x0)
g(x0)
(9) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x) → G(f(x))
F(x) → F(x)
The TRS R consists of the following rules:
f(x) → g(f(x))
g(x) → a
The set Q consists of the following terms:
f(x0)
g(x0)
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x) → F(x)
The TRS R consists of the following rules:
f(x) → g(f(x))
g(x) → a
The set Q consists of the following terms:
f(x0)
g(x0)
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x) → F(x)
R is empty.
The set Q consists of the following terms:
f(x0)
g(x0)
We have to consider all minimal (P,Q,R)-chains.
(15) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(x0)
g(x0)
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x) → F(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x) → F(x)
R is empty.
The set Q consists of the following terms:
f(x0)
g(x0)
We have to consider all minimal (P,Q,R)-chains.
(19) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(x0)
g(x0)
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x) → F(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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 =
F(
x) evaluates to t =
F(
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 F(x) to F(x).
(22) FALSE