(0) Obligation:
Clauses:
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
Query: app(g,a,a)
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph ICLP10.
(2) Obligation:
Clauses:
appA([], T5, T5).
appA(.(T10, []), T20, .(T10, T20)).
appA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) :- appA(T30, T33, T34).
Query: appA(g,a,a)
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
appA_in: (b,f,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
appA_in_gaa([], T5, T5) → appA_out_gaa([], T5, T5)
appA_in_gaa(.(T10, []), T20, .(T10, T20)) → appA_out_gaa(.(T10, []), T20, .(T10, T20))
appA_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, appA_out_gaa(T30, T33, T34)) → appA_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
[] =
[]
appA_out_gaa(
x1,
x2,
x3) =
appA_out_gaa
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gaa(
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:
appA_in_gaa([], T5, T5) → appA_out_gaa([], T5, T5)
appA_in_gaa(.(T10, []), T20, .(T10, T20)) → appA_out_gaa(.(T10, []), T20, .(T10, T20))
appA_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, appA_out_gaa(T30, T33, T34)) → appA_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
[] =
[]
appA_out_gaa(
x1,
x2,
x3) =
appA_out_gaa
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gaa(
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:
APPA_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_GAA(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
APPA_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APPA_IN_GAA(T30, T33, T34)
The TRS R consists of the following rules:
appA_in_gaa([], T5, T5) → appA_out_gaa([], T5, T5)
appA_in_gaa(.(T10, []), T20, .(T10, T20)) → appA_out_gaa(.(T10, []), T20, .(T10, T20))
appA_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, appA_out_gaa(T30, T33, T34)) → appA_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
[] =
[]
appA_out_gaa(
x1,
x2,
x3) =
appA_out_gaa
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gaa(
x6)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GAA(
x6)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APPA_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_GAA(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
APPA_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APPA_IN_GAA(T30, T33, T34)
The TRS R consists of the following rules:
appA_in_gaa([], T5, T5) → appA_out_gaa([], T5, T5)
appA_in_gaa(.(T10, []), T20, .(T10, T20)) → appA_out_gaa(.(T10, []), T20, .(T10, T20))
appA_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, appA_out_gaa(T30, T33, T34)) → appA_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
[] =
[]
appA_out_gaa(
x1,
x2,
x3) =
appA_out_gaa
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gaa(
x6)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GAA(
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:
APPA_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APPA_IN_GAA(T30, T33, T34)
The TRS R consists of the following rules:
appA_in_gaa([], T5, T5) → appA_out_gaa([], T5, T5)
appA_in_gaa(.(T10, []), T20, .(T10, T20)) → appA_out_gaa(.(T10, []), T20, .(T10, T20))
appA_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, appA_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, appA_out_gaa(T30, T33, T34)) → appA_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
[] =
[]
appA_out_gaa(
x1,
x2,
x3) =
appA_out_gaa
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gaa(
x6)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
x1)
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:
APPA_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APPA_IN_GAA(T30, T33, T34)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
x1)
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:
APPA_IN_GAA(.(T10, .(T29, T30))) → APPA_IN_GAA(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:
- APPA_IN_GAA(.(T10, .(T29, T30))) → APPA_IN_GAA(T30)
The graph contains the following edges 1 > 1
(14) YES