(0) Obligation:
Clauses:
p(0).
p(s(X)) :- ','(q(X), ','(!, r)).
p(X) :- r.
q(0).
q(s(X)) :- ','(p(X), ','(!, r)).
q(X) :- r.
r.
Query: p(g)
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph ICLP10.
(2) Obligation:
Clauses:
pA(0).
pA(0).
pA(s(0)).
pA(s(s(T10))) :- pA(T10).
pA(s(s(T10))) :- pA(T10).
pA(s(T13)).
pA(s(T16)).
pA(T18).
Query: pA(g)
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
pA_in: (b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))
The argument filtering Pi contains the following mapping:
pA_in_g(
x1) =
pA_in_g(
x1)
0 =
0
pA_out_g(
x1) =
pA_out_g
s(
x1) =
s(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))
The argument filtering Pi contains the following mapping:
pA_in_g(
x1) =
pA_in_g(
x1)
0 =
0
pA_out_g(
x1) =
pA_out_g
s(
x1) =
s(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
PA_IN_G(s(s(T10))) → U1_G(T10, pA_in_g(T10))
PA_IN_G(s(s(T10))) → PA_IN_G(T10)
The TRS R consists of the following rules:
pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))
The argument filtering Pi contains the following mapping:
pA_in_g(
x1) =
pA_in_g(
x1)
0 =
0
pA_out_g(
x1) =
pA_out_g
s(
x1) =
s(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
PA_IN_G(
x1) =
PA_IN_G(
x1)
U1_G(
x1,
x2) =
U1_G(
x2)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
PA_IN_G(s(s(T10))) → U1_G(T10, pA_in_g(T10))
PA_IN_G(s(s(T10))) → PA_IN_G(T10)
The TRS R consists of the following rules:
pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))
The argument filtering Pi contains the following mapping:
pA_in_g(
x1) =
pA_in_g(
x1)
0 =
0
pA_out_g(
x1) =
pA_out_g
s(
x1) =
s(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
PA_IN_G(
x1) =
PA_IN_G(
x1)
U1_G(
x1,
x2) =
U1_G(
x2)
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
PA_IN_G(s(s(T10))) → PA_IN_G(T10)
The TRS R consists of the following rules:
pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))
The argument filtering Pi contains the following mapping:
pA_in_g(
x1) =
pA_in_g(
x1)
0 =
0
pA_out_g(
x1) =
pA_out_g
s(
x1) =
s(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
PA_IN_G(
x1) =
PA_IN_G(
x1)
We have to consider all (P,R,Pi)-chains
(9) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(10) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
PA_IN_G(s(s(T10))) → PA_IN_G(T10)
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(11) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PA_IN_G(s(s(T10))) → PA_IN_G(T10)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) 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:
- PA_IN_G(s(s(T10))) → PA_IN_G(T10)
The graph contains the following edges 1 > 1
(14) YES