(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f → f
g(b) → c
b → c
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'(x)
b'(g(x)) → c'(x)
b'(x) → c'(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'(x)
b'(g(x)) → c'(x)
b'(x) → c'(x)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(b) = 1
POL(c) = 0
POL(f) = 0
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(b) → c
b → c
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f → f
Q is empty.
(7) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f → f
The signature Sigma is {
f}
(8) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f → f
The set Q consists of the following terms:
f
(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 → F
The TRS R consists of the following rules:
f → f
The set Q consists of the following terms:
f
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:
F → F
R is empty.
The set Q consists of the following terms:
f
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].
f
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F → F
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:
F → F
R is empty.
The set Q consists of the following terms:
f
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].
f
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F → F
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 =
F evaluates to t =
FThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from F to F.
(20) FALSE