(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c → f(g(c))
f(g(X)) → 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:
c'(x) → c'(g(f(x)))
g(f(x)) → g(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:
c'(x) → c'(g(f(x)))
g(f(x)) → g(x)
Q is empty.
(5) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c → f(g(c))
f(g(X)) → g(X)
The set Q consists of the following terms:
c
f(g(x0))
(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:
C → F(g(c))
C → C
The TRS R consists of the following rules:
c → f(g(c))
f(g(X)) → g(X)
The set Q consists of the following terms:
c
f(g(x0))
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:
C → C
The TRS R consists of the following rules:
c → f(g(c))
f(g(X)) → g(X)
The set Q consists of the following terms:
c
f(g(x0))
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C → C
R is empty.
The set Q consists of the following terms:
c
f(g(x0))
We have to consider all minimal (P,Q,R)-chains.
(13) 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].
c
f(g(x0))
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C → C
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C → C
R is empty.
The set Q consists of the following terms:
c
f(g(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) 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].
c
f(g(x0))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C → C
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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 =
C evaluates to t =
CThus 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 C to C.
(20) FALSE