(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(tt, x) → f(isDouble(x), s(s(x)))
isDouble(s(s(x))) → isDouble(x)
isDouble(0) → tt
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(tt, x) → F(isDouble(x), s(s(x)))
F(tt, x) → ISDOUBLE(x)
ISDOUBLE(s(s(x))) → ISDOUBLE(x)
The TRS R consists of the following rules:
f(tt, x) → f(isDouble(x), s(s(x)))
isDouble(s(s(x))) → isDouble(x)
isDouble(0) → tt
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:
ISDOUBLE(s(s(x))) → ISDOUBLE(x)
The TRS R consists of the following rules:
f(tt, x) → f(isDouble(x), s(s(x)))
isDouble(s(s(x))) → isDouble(x)
isDouble(0) → tt
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:
- ISDOUBLE(s(s(x))) → ISDOUBLE(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(tt, x) → F(isDouble(x), s(s(x)))
The TRS R consists of the following rules:
f(tt, x) → f(isDouble(x), s(s(x)))
isDouble(s(s(x))) → isDouble(x)
isDouble(0) → tt
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 = 1,
σ' = [ ], and μ' = [ ] on the rule
F(tt, s(s(zr0)))[zr0 / s(s(zr0))]n[zr0 / 0] → F(tt, s(s(s(s(zr0)))))[zr0 / s(s(zr0))]n[zr0 / 0]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(tt, s(s(zl1)))[zl1 / s(s(zl1)), zr1 / s(s(zr1))]n[zl1 / 0, zr1 / 0] → F(tt, s(s(s(s(zr1)))))[zl1 / s(s(zl1)), zr1 / s(s(zr1))]n[zl1 / 0, zr1 / 0]
by Narrowing at position: [0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
F(tt, s(s(zs1)))[zs1 / s(s(zs1))]n[zs1 / y0] → F(isDouble(y0), s(s(s(s(zs1)))))[zs1 / s(s(zs1))]n[zs1 / y0]
by Narrowing at position: [0]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
F(tt, x)[ ]n[ ] → F(isDouble(x), s(s(x)))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
isDouble(s(s(x)))[x / s(s(x))]n[ ] → isDouble(x)[ ]n[ ]
by PatternCreation I
isDouble(s(s(x)))[ ]n[ ] → isDouble(x)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
isDouble(0)[ ]n[ ] → tt[ ]n[ ]
by OriginalRule from TRS R
(10) NO