(0) Obligation:
Clauses:interleave([], Xs, Xs).
interleave(.(X, Xs), Ys, .(X, Zs)) :- interleave(Ys, Xs, Zs).
Queries:
interleave(g,g,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:interleave1([], T5, T5).
interleave1(.(T10, T19), [], .(T10, T19)).
interleave1(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) :- interleave1(T30, T29, T32).
Queries:
interleave1(g,g,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:interleave1_in: (b,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
interleave1_in_gga([], T5, T5) → interleave1_out_gga([], T5, T5)
interleave1_in_gga(.(T10, T19), [], .(T10, T19)) → interleave1_out_gga(.(T10, T19), [], .(T10, T19))
interleave1_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleave1_out_gga(T30, T29, T32)) → interleave1_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3) = interleave1_in_gga(x1, x2)
[] = []
interleave1_out_gga(x1, x2, x3) = interleave1_out_gga(x3)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6) = U1_gga(x1, x3, x6)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:The TRS R consists of the following rules:
interleave1_in_gga([], T5, T5) → interleave1_out_gga([], T5, T5)
interleave1_in_gga(.(T10, T19), [], .(T10, T19)) → interleave1_out_gga(.(T10, T19), [], .(T10, T19))
interleave1_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleave1_out_gga(T30, T29, T32)) → interleave1_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3) = interleave1_in_gga(x1, x2)
[] = []
interleave1_out_gga(x1, x2, x3) = interleave1_out_gga(x3)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6) = U1_gga(x1, x3, x6)
(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:
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_GGA(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)
The TRS R consists of the following rules:
interleave1_in_gga([], T5, T5) → interleave1_out_gga([], T5, T5)
interleave1_in_gga(.(T10, T19), [], .(T10, T19)) → interleave1_out_gga(.(T10, T19), [], .(T10, T19))
interleave1_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleave1_out_gga(T30, T29, T32)) → interleave1_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3) = interleave1_in_gga(x1, x2)
[] = []
interleave1_out_gga(x1, x2, x3) = interleave1_out_gga(x3)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6) = U1_gga(x1, x3, x6)
INTERLEAVE1_IN_GGA(x1, x2, x3) = INTERLEAVE1_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6) = U1_GGA(x1, x3, x6)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_GGA(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)
The TRS R consists of the following rules:
interleave1_in_gga([], T5, T5) → interleave1_out_gga([], T5, T5)
interleave1_in_gga(.(T10, T19), [], .(T10, T19)) → interleave1_out_gga(.(T10, T19), [], .(T10, T19))
interleave1_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleave1_out_gga(T30, T29, T32)) → interleave1_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3) = interleave1_in_gga(x1, x2)
[] = []
interleave1_out_gga(x1, x2, x3) = interleave1_out_gga(x3)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6) = U1_gga(x1, x3, x6)
INTERLEAVE1_IN_GGA(x1, x2, x3) = INTERLEAVE1_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6) = U1_GGA(x1, x3, x6)
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:
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)
The TRS R consists of the following rules:
interleave1_in_gga([], T5, T5) → interleave1_out_gga([], T5, T5)
interleave1_in_gga(.(T10, T19), [], .(T10, T19)) → interleave1_out_gga(.(T10, T19), [], .(T10, T19))
interleave1_in_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_gga(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
U1_gga(T10, T30, T28, T29, T32, interleave1_out_gga(T30, T29, T32)) → interleave1_out_gga(.(T10, T30), .(T28, T29), .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3) = interleave1_in_gga(x1, x2)
[] = []
interleave1_out_gga(x1, x2, x3) = interleave1_out_gga(x3)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5, x6) = U1_gga(x1, x3, x6)
INTERLEAVE1_IN_GGA(x1, x2, x3) = INTERLEAVE1_IN_GGA(x1, x2)
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:
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
INTERLEAVE1_IN_GGA(x1, x2, x3) = INTERLEAVE1_IN_GGA(x1, x2)
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:
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29)) → INTERLEAVE1_IN_GGA(T30, T29)
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:
- INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29)) → INTERLEAVE1_IN_GGA(T30, T29)
The graph contains the following edges 1 > 1, 2 > 2