(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(tt, x) → f(isList(x), Cons(tt, x))
isList(Cons(tt, xs)) → isList(xs)
isList(nil) → 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(isList(x), Cons(tt, x))
F(tt, x) → ISLIST(x)
ISLIST(Cons(tt, xs)) → ISLIST(xs)
The TRS R consists of the following rules:
f(tt, x) → f(isList(x), Cons(tt, x))
isList(Cons(tt, xs)) → isList(xs)
isList(nil) → 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:
ISLIST(Cons(tt, xs)) → ISLIST(xs)
The TRS R consists of the following rules:
f(tt, x) → f(isList(x), Cons(tt, x))
isList(Cons(tt, xs)) → isList(xs)
isList(nil) → 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:
- ISLIST(Cons(tt, xs)) → ISLIST(xs)
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(isList(x), Cons(tt, x))
The TRS R consists of the following rules:
f(tt, x) → f(isList(x), Cons(tt, x))
isList(Cons(tt, xs)) → isList(xs)
isList(nil) → 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, Cons(tt, zr0))[zr0 / Cons(tt, zr0)]n[zr0 / nil] → F(tt, Cons(tt, Cons(tt, zr0)))[zr0 / Cons(tt, zr0)]n[zr0 / nil]
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, Cons(tt, zl1))[zl1 / Cons(tt, zl1), zr1 / Cons(tt, zr1)]n[zl1 / nil, zr1 / nil] → F(tt, Cons(tt, Cons(tt, zr1)))[zl1 / Cons(tt, zl1), zr1 / Cons(tt, zr1)]n[zl1 / nil, zr1 / nil]
by Narrowing at position: [0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
F(tt, Cons(tt, zs1))[zs1 / Cons(tt, zs1)]n[zs1 / y0] → F(isList(y0), Cons(tt, Cons(tt, zs1)))[zs1 / Cons(tt, zs1)]n[zs1 / y0]
by Narrowing at position: [0]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
F(tt, x)[ ]n[ ] → F(isList(x), Cons(tt, x))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
isList(Cons(tt, xs))[xs / Cons(tt, xs)]n[ ] → isList(xs)[ ]n[ ]
by PatternCreation I
isList(Cons(tt, xs))[ ]n[ ] → isList(xs)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
isList(nil)[ ]n[ ] → tt[ ]n[ ]
by OriginalRule from TRS R
(10) NO