(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → f(c)
c → b
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) → c'(f(x))
c'(x) → b'(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) → c'(f(x))
c'(x) → b'(x)
Q is empty.
(5) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
c → b
The TRS R 2 is
f(X) → f(c)
The signature Sigma is {
f}
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → f(c)
c → b
The set Q consists of the following terms:
f(x0)
c
(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:
F(X) → F(c)
F(X) → C
The TRS R consists of the following rules:
f(X) → f(c)
c → b
The set Q consists of the following terms:
f(x0)
c
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:
F(X) → F(c)
The TRS R consists of the following rules:
f(X) → f(c)
c → b
The set Q consists of the following terms:
f(x0)
c
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(X) → F(c)
The TRS R consists of the following rules:
c → b
The set Q consists of the following terms:
f(x0)
c
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(x0)
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(c)
The TRS R consists of the following rules:
c → b
The set Q consists of the following terms:
c
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(X) → F(c)
The TRS R consists of the following rules:
c → b
The set Q consists of the following terms:
f(x0)
c
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(x0)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(c)
The TRS R consists of the following rules:
c → b
The set Q consists of the following terms:
c
We have to consider all minimal (P,Q,R)-chains.
(19) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(c)
The TRS R consists of the following rules:
c → b
Q is empty.
We have to consider all (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(
c)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [X / c]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from F(X) to F(c).
(22) FALSE