(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
a → 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:
h(x) → g(x)
a'(g(x)) → b'(f(x))
f(x) → a'(h(x))
a'(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:
h(x) → g(x)
a'(g(x)) → b'(f(x))
f(x) → a'(h(x))
a'(x) → b'(x)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a) = 1
POL(b) = 0
POL(f(x1)) = 1 + 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:
a → b
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
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
h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
The signature Sigma is {
f,
g,
h}
(8) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
The set Q consists of the following terms:
h(x0)
g(a)
f(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:
H(X) → G(X)
G(a) → F(b)
F(X) → H(a)
The TRS R consists of the following rules:
h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
The set Q consists of the following terms:
h(x0)
g(a)
f(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:
H(X) → G(X)
G(a) → F(b)
F(X) → H(a)
R is empty.
The set Q consists of the following terms:
h(x0)
g(a)
f(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].
h(x0)
g(a)
f(x0)
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(X) → G(X)
G(a) → F(b)
F(X) → H(a)
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:
H(X) → G(X)
G(a) → F(b)
F(X) → H(a)
R is empty.
The set Q consists of the following terms:
h(x0)
g(a)
f(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].
h(x0)
g(a)
f(x0)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(X) → G(X)
G(a) → F(b)
F(X) → H(a)
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 narrowing to the left:
s =
G(
a) evaluates to t =
G(
a)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequenceG(a) →
F(
b)
with rule
G(
a) →
F(
b) at position [] and matcher [ ]
F(b) →
H(
a)
with rule
F(
X') →
H(
a) at position [] and matcher [
X' /
b]
H(a) →
G(
a)
with rule
H(
X) →
G(
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.
(20) FALSE