(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(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(g(x), x, y) → F(y, y, g(y))
F(g(x), x, y) → G(y)
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(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 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(g(x), x, y) → F(y, y, g(y))
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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' = [x1 / g(x0)] on the rule
F(g(g(x0)), g(x0), g(g(x0)))[ ]n[ ] → F(g(g(x0)), g(x0), g(g(x0)))[ ]n[x1 / g(x0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
F(g(x1), x1, g(g(y0)))[ ]n[ ] → F(g(g(y0)), g(y0), g(g(y0)))[ ]n[ ]
by Narrowing at position: [1]
intermediate steps: Instantiation - Instantiation
F(g(x1), x1, g(y0))[ ]n[ ] → F(g(y0), g(y0), g(y0))[ ]n[ ]
by Narrowing at position: [2]
intermediate steps: Instantiation - Instantiation
F(g(x), x, y)[ ]n[ ] → F(y, y, g(y))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Instantiation
g(g(x))[ ]n[ ] → g(x)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Instantiation
g(g(x))[ ]n[ ] → g(x)[ ]n[ ]
by OriginalRule from TRS R
(6) NO