(0) Obligation:
Clauses:
div(X, 0, Z, R) :- ','(!, fail).
div(0, Y, Z, R) :- ','(!, ','(=(Z, 0), =(R, 0))).
div(X, Y, s(Z), R) :- ','(minus(X, Y, U), ','(!, div(U, Y, Z, R))).
div(X, Y, 0, X).
minus(X, 0, X).
minus(s(X), s(Y), Z) :- minus(X, Y, Z).
Query: div(g,g,a,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
minusA(s(X1), s(X2), X3) :- minusA(X1, X2, X3).
divB(s(X1), s(X2), s(X3), X4) :- minusA(X1, X2, X5).
divB(s(X1), s(X2), s(X3), X4) :- ','(minuscA(X1, X2, X5), divB(X5, s(X2), X3, X4)).
Clauses:
minuscA(X1, 0, X1).
minuscA(s(X1), s(X2), X3) :- minuscA(X1, X2, X3).
divcB(0, X1, 0, 0).
divcB(s(X1), s(X2), s(X3), X4) :- ','(minuscA(X1, X2, X5), divcB(X5, s(X2), X3, X4)).
divcB(X1, X2, 0, X1).
divcB(X1, X2, 0, X1).
Afs:
divB(x1, x2, x3, x4) = divB(x1, x2)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
divB_in: (b,b,f,f)
minusA_in: (b,b,f)
minuscA_in: (b,b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U2_GGAA(X1, X2, X3, X4, minusA_in_gga(X1, X2, X5))
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → MINUSA_IN_GGA(X1, X2, X5)
MINUSA_IN_GGA(s(X1), s(X2), X3) → U1_GGA(X1, X2, X3, minusA_in_gga(X1, X2, X3))
MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U3_GGAA(X1, X2, X3, X4, minuscA_in_gga(X1, X2, X5))
U3_GGAA(X1, X2, X3, X4, minuscA_out_gga(X1, X2, X5)) → U4_GGAA(X1, X2, X3, X4, divB_in_ggaa(X5, s(X2), X3, X4))
U3_GGAA(X1, X2, X3, X4, minuscA_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, s(X2), X3, X4)
The TRS R consists of the following rules:
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
divB_in_ggaa(
x1,
x2,
x3,
x4) =
divB_in_ggaa(
x1,
x2)
s(
x1) =
s(
x1)
minusA_in_gga(
x1,
x2,
x3) =
minusA_in_gga(
x1,
x2)
minuscA_in_gga(
x1,
x2,
x3) =
minuscA_in_gga(
x1,
x2)
0 =
0
minuscA_out_gga(
x1,
x2,
x3) =
minuscA_out_gga(
x1,
x2,
x3)
U6_gga(
x1,
x2,
x3,
x4) =
U6_gga(
x1,
x2,
x4)
DIVB_IN_GGAA(
x1,
x2,
x3,
x4) =
DIVB_IN_GGAA(
x1,
x2)
U2_GGAA(
x1,
x2,
x3,
x4,
x5) =
U2_GGAA(
x1,
x2,
x5)
MINUSA_IN_GGA(
x1,
x2,
x3) =
MINUSA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
U3_GGAA(
x1,
x2,
x3,
x4,
x5) =
U3_GGAA(
x1,
x2,
x5)
U4_GGAA(
x1,
x2,
x3,
x4,
x5) =
U4_GGAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U2_GGAA(X1, X2, X3, X4, minusA_in_gga(X1, X2, X5))
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → MINUSA_IN_GGA(X1, X2, X5)
MINUSA_IN_GGA(s(X1), s(X2), X3) → U1_GGA(X1, X2, X3, minusA_in_gga(X1, X2, X3))
MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U3_GGAA(X1, X2, X3, X4, minuscA_in_gga(X1, X2, X5))
U3_GGAA(X1, X2, X3, X4, minuscA_out_gga(X1, X2, X5)) → U4_GGAA(X1, X2, X3, X4, divB_in_ggaa(X5, s(X2), X3, X4))
U3_GGAA(X1, X2, X3, X4, minuscA_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, s(X2), X3, X4)
The TRS R consists of the following rules:
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
divB_in_ggaa(
x1,
x2,
x3,
x4) =
divB_in_ggaa(
x1,
x2)
s(
x1) =
s(
x1)
minusA_in_gga(
x1,
x2,
x3) =
minusA_in_gga(
x1,
x2)
minuscA_in_gga(
x1,
x2,
x3) =
minuscA_in_gga(
x1,
x2)
0 =
0
minuscA_out_gga(
x1,
x2,
x3) =
minuscA_out_gga(
x1,
x2,
x3)
U6_gga(
x1,
x2,
x3,
x4) =
U6_gga(
x1,
x2,
x4)
DIVB_IN_GGAA(
x1,
x2,
x3,
x4) =
DIVB_IN_GGAA(
x1,
x2)
U2_GGAA(
x1,
x2,
x3,
x4,
x5) =
U2_GGAA(
x1,
x2,
x5)
MINUSA_IN_GGA(
x1,
x2,
x3) =
MINUSA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
U3_GGAA(
x1,
x2,
x3,
x4,
x5) =
U3_GGAA(
x1,
x2,
x5)
U4_GGAA(
x1,
x2,
x3,
x4,
x5) =
U4_GGAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)
The TRS R consists of the following rules:
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
minuscA_in_gga(
x1,
x2,
x3) =
minuscA_in_gga(
x1,
x2)
0 =
0
minuscA_out_gga(
x1,
x2,
x3) =
minuscA_out_gga(
x1,
x2,
x3)
U6_gga(
x1,
x2,
x3,
x4) =
U6_gga(
x1,
x2,
x4)
MINUSA_IN_GGA(
x1,
x2,
x3) =
MINUSA_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(8) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(9) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
MINUSA_IN_GGA(
x1,
x2,
x3) =
MINUSA_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(10) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUSA_IN_GGA(s(X1), s(X2)) → MINUSA_IN_GGA(X1, X2)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) 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:
- MINUSA_IN_GGA(s(X1), s(X2)) → MINUSA_IN_GGA(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2
(13) YES
(14) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U3_GGAA(X1, X2, X3, X4, minuscA_in_gga(X1, X2, X5))
U3_GGAA(X1, X2, X3, X4, minuscA_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, s(X2), X3, X4)
The TRS R consists of the following rules:
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
minuscA_in_gga(
x1,
x2,
x3) =
minuscA_in_gga(
x1,
x2)
0 =
0
minuscA_out_gga(
x1,
x2,
x3) =
minuscA_out_gga(
x1,
x2,
x3)
U6_gga(
x1,
x2,
x3,
x4) =
U6_gga(
x1,
x2,
x4)
DIVB_IN_GGAA(
x1,
x2,
x3,
x4) =
DIVB_IN_GGAA(
x1,
x2)
U3_GGAA(
x1,
x2,
x3,
x4,
x5) =
U3_GGAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
(15) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DIVB_IN_GGAA(s(X1), s(X2)) → U3_GGAA(X1, X2, minuscA_in_gga(X1, X2))
U3_GGAA(X1, X2, minuscA_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, s(X2))
The TRS R consists of the following rules:
minuscA_in_gga(X1, 0) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2)) → U6_gga(X1, X2, minuscA_in_gga(X1, X2))
U6_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
The set Q consists of the following terms:
minuscA_in_gga(x0, x1)
U6_gga(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(17) QDPQMonotonicMRRProof (EQUIVALENT transformation)
By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.
Strictly oriented dependency pairs:
DIVB_IN_GGAA(s(X1), s(X2)) → U3_GGAA(X1, X2, minuscA_in_gga(X1, X2))
Strictly oriented rules of the TRS R:
U6_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(DIVB_IN_GGAA(x1, x2)) = 2·x1
POL(U3_GGAA(x1, x2, x3)) = 2·x3
POL(U6_gga(x1, x2, x3)) = 2 + 2·x1 + x3
POL(minuscA_in_gga(x1, x2)) = 2·x1
POL(minuscA_out_gga(x1, x2, x3)) = x3
POL(s(x1)) = 1 + 2·x1
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U3_GGAA(X1, X2, minuscA_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, s(X2))
The TRS R consists of the following rules:
minuscA_in_gga(X1, 0) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2)) → U6_gga(X1, X2, minuscA_in_gga(X1, X2))
The set Q consists of the following terms:
minuscA_in_gga(x0, x1)
U6_gga(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(20) TRUE