(0) Obligation:
Clauses:
p(s(X), X).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).
Queries:
plus(g,a,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:
plus1(0, T5, T5).
plus1(s(T18), T12, s(T13)) :- plus1(T18, T12, T13).
Queries:
plus1(g,a,a).
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus1_in: (b,f,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))
The argument filtering Pi contains the following mapping:
plus1_in_gaa(
x1,
x2,
x3) =
plus1_in_gaa(
x1)
0 =
0
plus1_out_gaa(
x1,
x2,
x3) =
plus1_out_gaa
s(
x1) =
s(
x1)
U1_gaa(
x1,
x2,
x3,
x4) =
U1_gaa(
x4)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))
The argument filtering Pi contains the following mapping:
plus1_in_gaa(
x1,
x2,
x3) =
plus1_in_gaa(
x1)
0 =
0
plus1_out_gaa(
x1,
x2,
x3) =
plus1_out_gaa
s(
x1) =
s(
x1)
U1_gaa(
x1,
x2,
x3,
x4) =
U1_gaa(
x4)
(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:
PLUS1_IN_GAA(s(T18), T12, s(T13)) → U1_GAA(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))
The argument filtering Pi contains the following mapping:
plus1_in_gaa(
x1,
x2,
x3) =
plus1_in_gaa(
x1)
0 =
0
plus1_out_gaa(
x1,
x2,
x3) =
plus1_out_gaa
s(
x1) =
s(
x1)
U1_gaa(
x1,
x2,
x3,
x4) =
U1_gaa(
x4)
PLUS1_IN_GAA(
x1,
x2,
x3) =
PLUS1_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4) =
U1_GAA(
x4)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
PLUS1_IN_GAA(s(T18), T12, s(T13)) → U1_GAA(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))
The argument filtering Pi contains the following mapping:
plus1_in_gaa(
x1,
x2,
x3) =
plus1_in_gaa(
x1)
0 =
0
plus1_out_gaa(
x1,
x2,
x3) =
plus1_out_gaa
s(
x1) =
s(
x1)
U1_gaa(
x1,
x2,
x3,
x4) =
U1_gaa(
x4)
PLUS1_IN_GAA(
x1,
x2,
x3) =
PLUS1_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4) =
U1_GAA(
x4)
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:
PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)
The TRS R consists of the following rules:
plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(T18), T12, s(T13)) → U1_gaa(T18, T12, T13, plus1_in_gaa(T18, T12, T13))
U1_gaa(T18, T12, T13, plus1_out_gaa(T18, T12, T13)) → plus1_out_gaa(s(T18), T12, s(T13))
The argument filtering Pi contains the following mapping:
plus1_in_gaa(
x1,
x2,
x3) =
plus1_in_gaa(
x1)
0 =
0
plus1_out_gaa(
x1,
x2,
x3) =
plus1_out_gaa
s(
x1) =
s(
x1)
U1_gaa(
x1,
x2,
x3,
x4) =
U1_gaa(
x4)
PLUS1_IN_GAA(
x1,
x2,
x3) =
PLUS1_IN_GAA(
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:
PLUS1_IN_GAA(s(T18), T12, s(T13)) → PLUS1_IN_GAA(T18, T12, T13)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
PLUS1_IN_GAA(
x1,
x2,
x3) =
PLUS1_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(11) PiDPToQDPProof (SOUND 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:
PLUS1_IN_GAA(s(T18)) → PLUS1_IN_GAA(T18)
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:
- PLUS1_IN_GAA(s(T18)) → PLUS1_IN_GAA(T18)
The graph contains the following edges 1 > 1
(14) YES