(0) Obligation:
Clauses:
avg(s(X), Y, Z) :- avg(X, s(Y), Z).
avg(X, s(s(s(Y))), s(Z)) :- avg(s(X), Y, Z).
avg(0, 0, 0).
avg(0, s(0), 0).
avg(0, s(s(0)), s(0)).
Query: avg(g,g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
avgA(s(s(X1)), X2, X3) :- avgA(X1, s(s(X2)), X3).
avgA(s(X1), s(s(X2)), s(X3)) :- avgA(s(X1), X2, X3).
avgA(s(X1), s(s(s(X2))), s(X3)) :- avgA(s(s(X1)), X2, X3).
avgA(X1, s(s(s(X2))), s(X3)) :- avgA(X1, s(X2), X3).
avgA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) :- avgA(s(s(X1)), X2, X3).
Clauses:
avgcA(s(s(X1)), X2, X3) :- avgcA(X1, s(s(X2)), X3).
avgcA(s(X1), s(s(X2)), s(X3)) :- avgcA(s(X1), X2, X3).
avgcA(s(0), 0, 0).
avgcA(s(0), s(0), s(0)).
avgcA(s(X1), s(s(s(X2))), s(X3)) :- avgcA(s(s(X1)), X2, X3).
avgcA(X1, s(s(s(X2))), s(X3)) :- avgcA(X1, s(X2), X3).
avgcA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) :- avgcA(s(s(X1)), X2, X3).
avgcA(0, 0, 0).
avgcA(0, s(0), 0).
avgcA(0, s(s(0)), s(0)).
Afs:
avgA(x1, x2, x3) = avgA(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:
avgA_in: (b,b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
AVGA_IN_GGA(s(s(X1)), X2, X3) → U1_GGA(X1, X2, X3, avgA_in_gga(X1, s(s(X2)), X3))
AVGA_IN_GGA(s(s(X1)), X2, X3) → AVGA_IN_GGA(X1, s(s(X2)), X3)
AVGA_IN_GGA(s(X1), s(s(X2)), s(X3)) → U2_GGA(X1, X2, X3, avgA_in_gga(s(X1), X2, X3))
AVGA_IN_GGA(s(X1), s(s(X2)), s(X3)) → AVGA_IN_GGA(s(X1), X2, X3)
AVGA_IN_GGA(s(X1), s(s(s(X2))), s(X3)) → U3_GGA(X1, X2, X3, avgA_in_gga(s(s(X1)), X2, X3))
AVGA_IN_GGA(s(X1), s(s(s(X2))), s(X3)) → AVGA_IN_GGA(s(s(X1)), X2, X3)
AVGA_IN_GGA(X1, s(s(s(X2))), s(X3)) → U4_GGA(X1, X2, X3, avgA_in_gga(X1, s(X2), X3))
AVGA_IN_GGA(X1, s(s(s(X2))), s(X3)) → AVGA_IN_GGA(X1, s(X2), X3)
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) → U5_GGA(X1, X2, X3, avgA_in_gga(s(s(X1)), X2, X3))
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) → AVGA_IN_GGA(s(s(X1)), X2, X3)
R is empty.
The argument filtering Pi contains the following mapping:
avgA_in_gga(
x1,
x2,
x3) =
avgA_in_gga(
x1,
x2)
s(
x1) =
s(
x1)
AVGA_IN_GGA(
x1,
x2,
x3) =
AVGA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
U2_GGA(
x1,
x2,
x3,
x4) =
U2_GGA(
x1,
x2,
x4)
U3_GGA(
x1,
x2,
x3,
x4) =
U3_GGA(
x1,
x2,
x4)
U4_GGA(
x1,
x2,
x3,
x4) =
U4_GGA(
x1,
x2,
x4)
U5_GGA(
x1,
x2,
x3,
x4) =
U5_GGA(
x1,
x2,
x4)
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:
AVGA_IN_GGA(s(s(X1)), X2, X3) → U1_GGA(X1, X2, X3, avgA_in_gga(X1, s(s(X2)), X3))
AVGA_IN_GGA(s(s(X1)), X2, X3) → AVGA_IN_GGA(X1, s(s(X2)), X3)
AVGA_IN_GGA(s(X1), s(s(X2)), s(X3)) → U2_GGA(X1, X2, X3, avgA_in_gga(s(X1), X2, X3))
AVGA_IN_GGA(s(X1), s(s(X2)), s(X3)) → AVGA_IN_GGA(s(X1), X2, X3)
AVGA_IN_GGA(s(X1), s(s(s(X2))), s(X3)) → U3_GGA(X1, X2, X3, avgA_in_gga(s(s(X1)), X2, X3))
AVGA_IN_GGA(s(X1), s(s(s(X2))), s(X3)) → AVGA_IN_GGA(s(s(X1)), X2, X3)
AVGA_IN_GGA(X1, s(s(s(X2))), s(X3)) → U4_GGA(X1, X2, X3, avgA_in_gga(X1, s(X2), X3))
AVGA_IN_GGA(X1, s(s(s(X2))), s(X3)) → AVGA_IN_GGA(X1, s(X2), X3)
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) → U5_GGA(X1, X2, X3, avgA_in_gga(s(s(X1)), X2, X3))
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) → AVGA_IN_GGA(s(s(X1)), X2, X3)
R is empty.
The argument filtering Pi contains the following mapping:
avgA_in_gga(
x1,
x2,
x3) =
avgA_in_gga(
x1,
x2)
s(
x1) =
s(
x1)
AVGA_IN_GGA(
x1,
x2,
x3) =
AVGA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
U2_GGA(
x1,
x2,
x3,
x4) =
U2_GGA(
x1,
x2,
x4)
U3_GGA(
x1,
x2,
x3,
x4) =
U3_GGA(
x1,
x2,
x4)
U4_GGA(
x1,
x2,
x3,
x4) =
U4_GGA(
x1,
x2,
x4)
U5_GGA(
x1,
x2,
x3,
x4) =
U5_GGA(
x1,
x2,
x4)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
AVGA_IN_GGA(s(X1), s(s(X2)), s(X3)) → AVGA_IN_GGA(s(X1), X2, X3)
AVGA_IN_GGA(s(s(X1)), X2, X3) → AVGA_IN_GGA(X1, s(s(X2)), X3)
AVGA_IN_GGA(s(X1), s(s(s(X2))), s(X3)) → AVGA_IN_GGA(s(s(X1)), X2, X3)
AVGA_IN_GGA(X1, s(s(s(X2))), s(X3)) → AVGA_IN_GGA(X1, s(X2), X3)
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2)))))), s(s(X3))) → AVGA_IN_GGA(s(s(X1)), X2, X3)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
AVGA_IN_GGA(
x1,
x2,
x3) =
AVGA_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AVGA_IN_GGA(s(X1), s(s(X2))) → AVGA_IN_GGA(s(X1), X2)
AVGA_IN_GGA(s(s(X1)), X2) → AVGA_IN_GGA(X1, s(s(X2)))
AVGA_IN_GGA(s(X1), s(s(s(X2)))) → AVGA_IN_GGA(s(s(X1)), X2)
AVGA_IN_GGA(X1, s(s(s(X2)))) → AVGA_IN_GGA(X1, s(X2))
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2))))))) → AVGA_IN_GGA(s(s(X1)), X2)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
AVGA_IN_GGA(s(X1), s(s(X2))) → AVGA_IN_GGA(s(X1), X2)
AVGA_IN_GGA(s(s(X1)), X2) → AVGA_IN_GGA(X1, s(s(X2)))
AVGA_IN_GGA(s(X1), s(s(s(X2)))) → AVGA_IN_GGA(s(s(X1)), X2)
AVGA_IN_GGA(X1, s(s(s(X2)))) → AVGA_IN_GGA(X1, s(X2))
AVGA_IN_GGA(X1, s(s(s(s(s(s(X2))))))) → AVGA_IN_GGA(s(s(X1)), X2)
Used ordering: Knuth-Bendix order [KBO] with precedence:
s1 > AVGAINGGA2
and weight map:
s_1=1
AVGA_IN_GGA_2=0
The variable weight is 1
(10) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(12) YES