(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
add(true, x, xs) → add(and(isNat(x), isList(xs)), x, Cons(x, xs))
isList(Cons(x, xs)) → isList(xs)
isList(nil) → true
isNat(s(x)) → isNat(x)
isNat(0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
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:
ADD(true, x, xs) → ADD(and(isNat(x), isList(xs)), x, Cons(x, xs))
ADD(true, x, xs) → AND(isNat(x), isList(xs))
ADD(true, x, xs) → ISNAT(x)
ADD(true, x, xs) → ISLIST(xs)
ISLIST(Cons(x, xs)) → ISLIST(xs)
ISNAT(s(x)) → ISNAT(x)
The TRS R consists of the following rules:
add(true, x, xs) → add(and(isNat(x), isList(xs)), x, Cons(x, xs))
isList(Cons(x, xs)) → isList(xs)
isList(nil) → true
isNat(s(x)) → isNat(x)
isNat(0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
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 3 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ISNAT(s(x)) → ISNAT(x)
The TRS R consists of the following rules:
add(true, x, xs) → add(and(isNat(x), isList(xs)), x, Cons(x, xs))
isList(Cons(x, xs)) → isList(xs)
isList(nil) → true
isNat(s(x)) → isNat(x)
isNat(0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
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:
- ISNAT(s(x)) → ISNAT(x)
The graph contains the following edges 1 > 1
(7) YES
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ISLIST(Cons(x, xs)) → ISLIST(xs)
The TRS R consists of the following rules:
add(true, x, xs) → add(and(isNat(x), isList(xs)), x, Cons(x, xs))
isList(Cons(x, xs)) → isList(xs)
isList(nil) → true
isNat(s(x)) → isNat(x)
isNat(0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) 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:
- ISLIST(Cons(x, xs)) → ISLIST(xs)
The graph contains the following edges 1 > 1
(10) YES
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(true, x, xs) → ADD(and(isNat(x), isList(xs)), x, Cons(x, xs))
The TRS R consists of the following rules:
add(true, x, xs) → add(and(isNat(x), isList(xs)), x, Cons(x, xs))
isList(Cons(x, xs)) → isList(xs)
isList(nil) → true
isNat(s(x)) → isNat(x)
isNat(0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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
ADD(true, 0, Cons(0, zr1))[zr1 / Cons(0, zr1)]n[zr1 / nil] → ADD(true, 0, Cons(0, Cons(0, zr1)))[zr1 / Cons(0, zr1)]n[zr1 / nil]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
ADD(true, 0, Cons(x0, zl1))[zl1 / Cons(x0, zl1), zr1 / Cons(x0, zr1)]n[zl1 / nil, zr1 / nil] → ADD(true, 0, Cons(0, Cons(x0, zr1)))[zl1 / Cons(x0, zl1), zr1 / Cons(x0, zr1)]n[zl1 / nil, zr1 / nil]
by Narrowing at position: [0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
ADD(true, 0, Cons(x0, zl1))[zl1 / Cons(x0, zl1), zr1 / Cons(x0, zr1)]n[zl1 / nil, zr1 / nil] → ADD(and(true, true), 0, Cons(0, Cons(x0, zr1)))[zl1 / Cons(x0, zl1), zr1 / Cons(x0, zr1)]n[zl1 / nil, zr1 / nil]
by Narrowing at position: [0,0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
ADD(true, x2, Cons(x1, zl1))[zl1 / Cons(x1, zl1), zr1 / Cons(x1, zr1)]n[zl1 / nil, zr1 / nil] → ADD(and(isNat(x2), true), x2, Cons(x2, Cons(x1, zr1)))[zl1 / Cons(x1, zl1), zr1 / Cons(x1, zr1)]n[zl1 / nil, zr1 / nil]
by Narrowing at position: [0,1]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
ADD(true, x1, Cons(y1, zs1))[zs1 / Cons(y1, zs1)]n[zs1 / y0] → ADD(and(isNat(x1), isList(y0)), x1, Cons(x1, Cons(y1, zs1)))[zs1 / Cons(y1, zs1)]n[zs1 / y0]
by Narrowing at position: [0,1]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
ADD(true, x, xs)[ ]n[ ] → ADD(and(isNat(x), isList(xs)), x, Cons(x, xs))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
isList(Cons(x, xs))[xs / Cons(x, xs)]n[ ] → isList(xs)[ ]n[ ]
by PatternCreation I
isList(Cons(x, xs))[ ]n[ ] → isList(xs)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
isList(nil)[ ]n[ ] → true[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
isNat(0)[ ]n[ ] → true[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
and(true, true)[ ]n[ ] → true[ ]n[ ]
by OriginalRule from TRS R
(13) NO