(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) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
isDouble(s(s(x))) → isDouble(x)
isDouble(0) → tt
The TRS R 2 is
f(tt, x) → f(isDouble(x), s(s(x)))
The signature Sigma is {
f}
(2) 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
The set Q consists of the following terms:
f(tt, x0)
isDouble(s(s(x0)))
isDouble(0)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) 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
The set Q consists of the following terms:
f(tt, x0)
isDouble(s(s(x0)))
isDouble(0)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) 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
The set Q consists of the following terms:
f(tt, x0)
isDouble(s(s(x0)))
isDouble(0)
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ISDOUBLE(s(s(x))) → ISDOUBLE(x)
R is empty.
The set Q consists of the following terms:
f(tt, x0)
isDouble(s(s(x0)))
isDouble(0)
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(tt, x0)
isDouble(s(s(x0)))
isDouble(0)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ISDOUBLE(s(s(x))) → ISDOUBLE(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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
(13) YES
(14) 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
The set Q consists of the following terms:
f(tt, x0)
isDouble(s(s(x0)))
isDouble(0)
We have to consider all minimal (P,Q,R)-chains.
(15) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
(16) 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 (P,Q,R)-chains.
(17) NonLoopProof (COMPLETE 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:
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]
by Equivalence by Domain Renaming of the lhs with [zl0 / zr0]
intermediate steps: Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
F(tt, s(s(zl1)))[zr1 / s(s(zr1)), zl1 / s(s(zl1))]n[zr1 / 0, zl1 / 0] → F(tt, s(s(s(s(zr1)))))[zr1 / s(s(zr1)), zl1 / s(s(zl1))]n[zr1 / 0, zl1 / 0]
by Narrowing at position: [0]
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
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 Rule from TRS P
intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
isDouble(s(s(x)))[x / s(s(x))]n[ ] → isDouble(x)[ ]n[ ]
by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(s(x))]
isDouble(s(s(x)))[ ]n[ ] → isDouble(x)[ ]n[ ]
by Rule from TRS R
intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
isDouble(0)[ ]n[ ] → tt[ ]n[ ]
by Rule from TRS R
(18) NO