(0) Obligation:
Clauses:
q(X) :- p(X, 0).
p(0, X1).
p(s(X), Y) :- p(X, s(Y)).
Query: q(g)
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
q_in: (b)
p_in: (b,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
q_in_g(X) → U1_g(X, p_in_gg(X, 0))
p_in_gg(0, X1) → p_out_gg(0, X1)
p_in_gg(s(X), Y) → U2_gg(X, Y, p_in_gg(X, s(Y)))
U2_gg(X, Y, p_out_gg(X, s(Y))) → p_out_gg(s(X), Y)
U1_g(X, p_out_gg(X, 0)) → q_out_g(X)
The argument filtering Pi contains the following mapping:
q_in_g(
x1) =
q_in_g(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
p_in_gg(
x1,
x2) =
p_in_gg(
x1,
x2)
0 =
0
p_out_gg(
x1,
x2) =
p_out_gg
s(
x1) =
s(
x1)
U2_gg(
x1,
x2,
x3) =
U2_gg(
x3)
q_out_g(
x1) =
q_out_g
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
q_in_g(X) → U1_g(X, p_in_gg(X, 0))
p_in_gg(0, X1) → p_out_gg(0, X1)
p_in_gg(s(X), Y) → U2_gg(X, Y, p_in_gg(X, s(Y)))
U2_gg(X, Y, p_out_gg(X, s(Y))) → p_out_gg(s(X), Y)
U1_g(X, p_out_gg(X, 0)) → q_out_g(X)
The argument filtering Pi contains the following mapping:
q_in_g(
x1) =
q_in_g(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
p_in_gg(
x1,
x2) =
p_in_gg(
x1,
x2)
0 =
0
p_out_gg(
x1,
x2) =
p_out_gg
s(
x1) =
s(
x1)
U2_gg(
x1,
x2,
x3) =
U2_gg(
x3)
q_out_g(
x1) =
q_out_g
(3) 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:
Q_IN_G(X) → U1_G(X, p_in_gg(X, 0))
Q_IN_G(X) → P_IN_GG(X, 0)
P_IN_GG(s(X), Y) → U2_GG(X, Y, p_in_gg(X, s(Y)))
P_IN_GG(s(X), Y) → P_IN_GG(X, s(Y))
The TRS R consists of the following rules:
q_in_g(X) → U1_g(X, p_in_gg(X, 0))
p_in_gg(0, X1) → p_out_gg(0, X1)
p_in_gg(s(X), Y) → U2_gg(X, Y, p_in_gg(X, s(Y)))
U2_gg(X, Y, p_out_gg(X, s(Y))) → p_out_gg(s(X), Y)
U1_g(X, p_out_gg(X, 0)) → q_out_g(X)
The argument filtering Pi contains the following mapping:
q_in_g(
x1) =
q_in_g(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
p_in_gg(
x1,
x2) =
p_in_gg(
x1,
x2)
0 =
0
p_out_gg(
x1,
x2) =
p_out_gg
s(
x1) =
s(
x1)
U2_gg(
x1,
x2,
x3) =
U2_gg(
x3)
q_out_g(
x1) =
q_out_g
Q_IN_G(
x1) =
Q_IN_G(
x1)
U1_G(
x1,
x2) =
U1_G(
x2)
P_IN_GG(
x1,
x2) =
P_IN_GG(
x1,
x2)
U2_GG(
x1,
x2,
x3) =
U2_GG(
x3)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
Q_IN_G(X) → U1_G(X, p_in_gg(X, 0))
Q_IN_G(X) → P_IN_GG(X, 0)
P_IN_GG(s(X), Y) → U2_GG(X, Y, p_in_gg(X, s(Y)))
P_IN_GG(s(X), Y) → P_IN_GG(X, s(Y))
The TRS R consists of the following rules:
q_in_g(X) → U1_g(X, p_in_gg(X, 0))
p_in_gg(0, X1) → p_out_gg(0, X1)
p_in_gg(s(X), Y) → U2_gg(X, Y, p_in_gg(X, s(Y)))
U2_gg(X, Y, p_out_gg(X, s(Y))) → p_out_gg(s(X), Y)
U1_g(X, p_out_gg(X, 0)) → q_out_g(X)
The argument filtering Pi contains the following mapping:
q_in_g(
x1) =
q_in_g(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
p_in_gg(
x1,
x2) =
p_in_gg(
x1,
x2)
0 =
0
p_out_gg(
x1,
x2) =
p_out_gg
s(
x1) =
s(
x1)
U2_gg(
x1,
x2,
x3) =
U2_gg(
x3)
q_out_g(
x1) =
q_out_g
Q_IN_G(
x1) =
Q_IN_G(
x1)
U1_G(
x1,
x2) =
U1_G(
x2)
P_IN_GG(
x1,
x2) =
P_IN_GG(
x1,
x2)
U2_GG(
x1,
x2,
x3) =
U2_GG(
x3)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_GG(s(X), Y) → P_IN_GG(X, s(Y))
The TRS R consists of the following rules:
q_in_g(X) → U1_g(X, p_in_gg(X, 0))
p_in_gg(0, X1) → p_out_gg(0, X1)
p_in_gg(s(X), Y) → U2_gg(X, Y, p_in_gg(X, s(Y)))
U2_gg(X, Y, p_out_gg(X, s(Y))) → p_out_gg(s(X), Y)
U1_g(X, p_out_gg(X, 0)) → q_out_g(X)
The argument filtering Pi contains the following mapping:
q_in_g(
x1) =
q_in_g(
x1)
U1_g(
x1,
x2) =
U1_g(
x2)
p_in_gg(
x1,
x2) =
p_in_gg(
x1,
x2)
0 =
0
p_out_gg(
x1,
x2) =
p_out_gg
s(
x1) =
s(
x1)
U2_gg(
x1,
x2,
x3) =
U2_gg(
x3)
q_out_g(
x1) =
q_out_g
P_IN_GG(
x1,
x2) =
P_IN_GG(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(7) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_GG(s(X), Y) → P_IN_GG(X, s(Y))
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(9) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P_IN_GG(s(X), Y) → P_IN_GG(X, s(Y))
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) 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:
- P_IN_GG(s(X), Y) → P_IN_GG(X, s(Y))
The graph contains the following edges 1 > 1
(12) YES