(0) Obligation:
Clauses:
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
Queries:
app(g,g,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:
app1([], T5, T5).
app1(.(T10, []), T19, .(T10, T19)).
app1(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) :- app1(T29, T30, T32).
Queries:
app1(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:
app1_in: (b,b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
app1_in_gga([], T5, T5) → app1_out_gga([], T5, T5)
app1_in_gga(.(T10, []), T19, .(T10, T19)) → app1_out_gga(.(T10, []), T19, .(T10, T19))
app1_in_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_gga(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
U1_gga(T10, T28, T29, T30, T32, app1_out_gga(T29, T30, T32)) → app1_out_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
app1_in_gga(
x1,
x2,
x3) =
app1_in_gga(
x1,
x2)
[] =
[]
app1_out_gga(
x1,
x2,
x3) =
app1_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
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:
app1_in_gga([], T5, T5) → app1_out_gga([], T5, T5)
app1_in_gga(.(T10, []), T19, .(T10, T19)) → app1_out_gga(.(T10, []), T19, .(T10, T19))
app1_in_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_gga(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
U1_gga(T10, T28, T29, T30, T32, app1_out_gga(T29, T30, T32)) → app1_out_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
app1_in_gga(
x1,
x2,
x3) =
app1_in_gga(
x1,
x2)
[] =
[]
app1_out_gga(
x1,
x2,
x3) =
app1_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
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:
APP1_IN_GGA(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_GGA(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
APP1_IN_GGA(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → APP1_IN_GGA(T29, T30, T32)
The TRS R consists of the following rules:
app1_in_gga([], T5, T5) → app1_out_gga([], T5, T5)
app1_in_gga(.(T10, []), T19, .(T10, T19)) → app1_out_gga(.(T10, []), T19, .(T10, T19))
app1_in_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_gga(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
U1_gga(T10, T28, T29, T30, T32, app1_out_gga(T29, T30, T32)) → app1_out_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
app1_in_gga(
x1,
x2,
x3) =
app1_in_gga(
x1,
x2)
[] =
[]
app1_out_gga(
x1,
x2,
x3) =
app1_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x6)
APP1_IN_GGA(
x1,
x2,
x3) =
APP1_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GGA(
x1,
x2,
x6)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APP1_IN_GGA(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_GGA(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
APP1_IN_GGA(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → APP1_IN_GGA(T29, T30, T32)
The TRS R consists of the following rules:
app1_in_gga([], T5, T5) → app1_out_gga([], T5, T5)
app1_in_gga(.(T10, []), T19, .(T10, T19)) → app1_out_gga(.(T10, []), T19, .(T10, T19))
app1_in_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_gga(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
U1_gga(T10, T28, T29, T30, T32, app1_out_gga(T29, T30, T32)) → app1_out_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
app1_in_gga(
x1,
x2,
x3) =
app1_in_gga(
x1,
x2)
[] =
[]
app1_out_gga(
x1,
x2,
x3) =
app1_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x6)
APP1_IN_GGA(
x1,
x2,
x3) =
APP1_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GGA(
x1,
x2,
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:
APP1_IN_GGA(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → APP1_IN_GGA(T29, T30, T32)
The TRS R consists of the following rules:
app1_in_gga([], T5, T5) → app1_out_gga([], T5, T5)
app1_in_gga(.(T10, []), T19, .(T10, T19)) → app1_out_gga(.(T10, []), T19, .(T10, T19))
app1_in_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → U1_gga(T10, T28, T29, T30, T32, app1_in_gga(T29, T30, T32))
U1_gga(T10, T28, T29, T30, T32, app1_out_gga(T29, T30, T32)) → app1_out_gga(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32)))
The argument filtering Pi contains the following mapping:
app1_in_gga(
x1,
x2,
x3) =
app1_in_gga(
x1,
x2)
[] =
[]
app1_out_gga(
x1,
x2,
x3) =
app1_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x6)
APP1_IN_GGA(
x1,
x2,
x3) =
APP1_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:
APP1_IN_GGA(.(T10, .(T28, T29)), T30, .(T10, .(T28, T32))) → APP1_IN_GGA(T29, T30, T32)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
APP1_IN_GGA(
x1,
x2,
x3) =
APP1_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:
APP1_IN_GGA(.(T10, .(T28, T29)), T30) → APP1_IN_GGA(T29, T30)
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:
- APP1_IN_GGA(.(T10, .(T28, T29)), T30) → APP1_IN_GGA(T29, T30)
The graph contains the following edges 1 > 1, 2 >= 2
(14) YES