(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
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:
F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
F(0, 1, g(x, y), z) → H(x)
H(g(x, y)) → H(x)
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
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 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(g(x, y)) → H(x)
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- H(g(x, y)) → H(x)
The graph contains the following edges 1 > 1
(7) YES
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
The TRS R consists of the following rules:
f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) NonLoopProof (EQUIVALENT transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x0 / h(0)] on the rule
F(0, 1, g(0, 1), h(0))[ ]n[ ] → F(0, 1, g(0, 1), h(0))[ ]n[x0 / h(0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu
F(0, 1, g(0, 1), x0)[ ]n[ ] → F(0, 1, g(0, 1), h(0))[ ]n[ ]
by Narrowing at position: [1]
F(0, 1, g(0, 1), x0)[ ]n[ ] → F(0, g(0, 1), g(0, 1), h(0))[ ]n[ ]
by Narrowing at position: [0]
intermediate steps: Instantiation - Instantiation - Instantiation
F(0, 1, g(x, y), z)[ ]n[ ] → F(g(x, y), g(x, y), g(x, y), h(x))[ ]n[ ]
by OriginalRule from TRS P
g(0, 1)[ ]n[ ] → 0[ ]n[ ]
by OriginalRule from TRS R
g(0, 1)[ ]n[ ] → 1[ ]n[ ]
by OriginalRule from TRS R
(10) NO