(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
APP(app(app(comp, f), g), x) → APP(g, x)
APP(twice, f) → APP(app(comp, f), f)
APP(twice, f) → APP(comp, f)
The TRS R consists of the following rules:
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(app(comp, f), g), x) → APP(g, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
The TRS R consists of the following rules:
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
APP(app(app(comp, f), g), x) → APP(g, x)
APP(app(app(comp, f), g), x) → APP(f, app(g, x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(
x1,
x2) =
APP(
x1)
app(
x1,
x2) =
app(
x1,
x2)
comp =
comp
twice =
twice
Lexicographic path order with status [LPO].
Quasi-Precedence:
comp > APP1 > [app2, twice]
Status:
APP1: [1]
app2: [1,2]
comp: []
twice: []
The following usable rules [FROCOS05] were oriented:
none
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app(app(app(comp, f), g), x) → app(f, app(g, x))
app(twice, f) → app(app(comp, f), f)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(8) TRUE