(0) Obligation:
Clauses:
len1([], 0).
len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M))).
eq(X, X).
Query: len1(g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
len1A(.(X1, X2), X3) :- len1A(X2, X4).
len1B(.(X1, .(X2, X3)), X4) :- len1A(X3, X5).
Clauses:
len1cA([], 0).
len1cA(.(X1, X2), s(X3)) :- len1cA(X2, X3).
Afs:
len1B(x1, x2) = len1B(x1)
(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:
len1B_in: (b,f)
len1A_in: (b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → U2_GA(X1, X2, X3, X4, len1A_in_ga(X3, X5))
LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → LEN1A_IN_GA(X3, X5)
LEN1A_IN_GA(.(X1, X2), X3) → U1_GA(X1, X2, X3, len1A_in_ga(X2, X4))
LEN1A_IN_GA(.(X1, X2), X3) → LEN1A_IN_GA(X2, X4)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
len1A_in_ga(
x1,
x2) =
len1A_in_ga(
x1)
LEN1B_IN_GA(
x1,
x2) =
LEN1B_IN_GA(
x1)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x3,
x5)
LEN1A_IN_GA(
x1,
x2) =
LEN1A_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
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:
LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → U2_GA(X1, X2, X3, X4, len1A_in_ga(X3, X5))
LEN1B_IN_GA(.(X1, .(X2, X3)), X4) → LEN1A_IN_GA(X3, X5)
LEN1A_IN_GA(.(X1, X2), X3) → U1_GA(X1, X2, X3, len1A_in_ga(X2, X4))
LEN1A_IN_GA(.(X1, X2), X3) → LEN1A_IN_GA(X2, X4)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
len1A_in_ga(
x1,
x2) =
len1A_in_ga(
x1)
LEN1B_IN_GA(
x1,
x2) =
LEN1B_IN_GA(
x1)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x3,
x5)
LEN1A_IN_GA(
x1,
x2) =
LEN1A_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
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 3 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LEN1A_IN_GA(.(X1, X2), X3) → LEN1A_IN_GA(X2, X4)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
LEN1A_IN_GA(
x1,
x2) =
LEN1A_IN_GA(
x1)
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:
LEN1A_IN_GA(.(X1, X2)) → LEN1A_IN_GA(X2)
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:
- LEN1A_IN_GA(.(X1, X2)) → LEN1A_IN_GA(X2)
The graph contains the following edges 1 > 1
(10) YES