(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0
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(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
F(s(s(s(s(s(s(s(s(x)))))))), y, y) → ID(s(s(s(s(s(s(s(s(x)))))))))
ID(s(x)) → ID(x)
The TRS R consists of the following rules:
f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0
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:
ID(s(x)) → ID(x)
The TRS R consists of the following rules:
f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0
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:
- ID(s(x)) → ID(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(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
The TRS R consists of the following rules:
f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0
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' = [zl0 / id(zl0)] on the rule
F(s(s(s(s(s(s(s(s(zl0)))))))), x1, x1)[zl0 / s(zl0)]n[ ] → F(s(s(s(s(s(s(s(s(zl0)))))))), x1, x1)[zl0 / s(zl0)]n[zl0 / id(zl0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Remove Unused) - Instantiate mu - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(s(s(s(s(s(s(s(s(x1)))))))), x0, x0)[x1 / s(x1), zt1 / s(zt1)]n[x1 / y0, zt1 / id(y0)] → F(s(s(s(s(s(s(s(s(zt1)))))))), x0, x0)[x1 / s(x1), zt1 / s(zt1)]n[x1 / y0, zt1 / id(y0)]
by Narrowing at position: [0]
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation
F(s(s(s(s(s(s(s(s(x)))))))), y, y)[ ]n[ ] → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Expand Sigma - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
id(s(x))[x / s(x)]n[ ] → s(z)[x / s(x), z / s(z)]n[z / id(x)]
by PatternCreation II
id(s(x))[ ]n[ ] → s(id(x))[ ]n[ ]
by OriginalRule from TRS R
(10) NO