(0) Obligation:
Clauses:
append1([], X, X).
append1(.(X, Y), U, .(X, Z)) :- append1(Y, U, Z).
append2([], X, X).
append2(.(X, Y), U, .(X, Z)) :- append2(Y, U, Z).
append3([], X, X).
append3(.(X, Y), U, .(X, Z)) :- append3(Y, U, Z).
Query: append3(a,a,g)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
append3A(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) :- append3A(X3, X4, X5).
append3A(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) :- append3A(X3, X4, X5).
Clauses:
append3cA([], X1, X1).
append3cA(.(X1, []), X2, .(X1, X2)).
append3cA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) :- append3cA(X3, X4, X5).
append3cA(.(X1, []), X2, .(X1, X2)).
append3cA(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) :- append3cA(X3, X4, X5).
Afs:
append3A(x1, x2, x3) = append3A(x3)
(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:
append3A_in: (f,f,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
APPEND3A_IN_AAG(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → U1_AAG(X1, X2, X3, X4, X5, append3A_in_aag(X3, X4, X5))
APPEND3A_IN_AAG(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → APPEND3A_IN_AAG(X3, X4, X5)
R is empty.
The argument filtering Pi contains the following mapping:
append3A_in_aag(
x1,
x2,
x3) =
append3A_in_aag(
x3)
.(
x1,
x2) =
.(
x1,
x2)
APPEND3A_IN_AAG(
x1,
x2,
x3) =
APPEND3A_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_AAG(
x1,
x2,
x5,
x6)
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:
APPEND3A_IN_AAG(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → U1_AAG(X1, X2, X3, X4, X5, append3A_in_aag(X3, X4, X5))
APPEND3A_IN_AAG(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → APPEND3A_IN_AAG(X3, X4, X5)
R is empty.
The argument filtering Pi contains the following mapping:
append3A_in_aag(
x1,
x2,
x3) =
append3A_in_aag(
x3)
.(
x1,
x2) =
.(
x1,
x2)
APPEND3A_IN_AAG(
x1,
x2,
x3) =
APPEND3A_IN_AAG(
x3)
U1_AAG(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_AAG(
x1,
x2,
x5,
x6)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APPEND3A_IN_AAG(.(X1, .(X2, X3)), X4, .(X1, .(X2, X5))) → APPEND3A_IN_AAG(X3, X4, X5)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
APPEND3A_IN_AAG(
x1,
x2,
x3) =
APPEND3A_IN_AAG(
x3)
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:
APPEND3A_IN_AAG(.(X1, .(X2, X5))) → APPEND3A_IN_AAG(X5)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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:
- APPEND3A_IN_AAG(.(X1, .(X2, X5))) → APPEND3A_IN_AAG(X5)
The graph contains the following edges 1 > 1
(10) YES