(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g → h
c → d
h → g
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:
g'(x) → h'(x)
c'(x) → d'(x)
h'(x) → g'(x)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(c'(x1)) = 1 + x1
POL(d'(x1)) = x1
POL(g'(x1)) = x1
POL(h'(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
c'(x) → d'(x)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g'(x) → h'(x)
h'(x) → g'(x)
Q is empty.
(5) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
g'(x) → h'(x)
h'(x) → g'(x)
The signature Sigma is {
g',
h'}
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
g'(x) → h'(x)
h'(x) → g'(x)
The set Q consists of the following terms:
g'(x0)
h'(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:
G'(x) → H'(x)
H'(x) → G'(x)
The TRS R consists of the following rules:
g'(x) → h'(x)
h'(x) → g'(x)
The set Q consists of the following terms:
g'(x0)
h'(x0)
We have to consider all minimal (P,Q,R)-chains.
(9) 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.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G'(x) → H'(x)
H'(x) → G'(x)
R is empty.
The set Q consists of the following terms:
g'(x0)
h'(x0)
We have to consider all minimal (P,Q,R)-chains.
(11) 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].
g'(x0)
h'(x0)
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G'(x) → H'(x)
H'(x) → G'(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
s =
H'(
x') evaluates to t =
H'(
x')
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceH'(x') →
G'(
x')
with rule
H'(
x'') →
G'(
x'') at position [] and matcher [
x'' /
x']
G'(x') →
H'(
x')
with rule
G'(
x) →
H'(
x)
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(14) NO