(0) Obligation:
Clauses:
reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).
Query: reverse(g,g,a)
(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)
Transformed Prolog program to (Pi-)TRS.
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverseA_in_gga([], T5, T5) → reverseA_out_gga([], T5, T5)
reverseA_in_gga(.(T28, []), .(T28, T29), T29) → reverseA_out_gga(.(T28, []), .(T28, T29), T29)
reverseA_in_gga(.(T43, .(T40, T41)), T42, T45) → U1_gga(T43, T40, T41, T42, T45, reverseA_in_gga(T41, T42, .(T40, .(T43, T45))))
U1_gga(T43, T40, T41, T42, T45, reverseA_out_gga(T41, T42, .(T40, .(T43, T45)))) → reverseA_out_gga(.(T43, .(T40, T41)), T42, T45)
The argument filtering Pi contains the following mapping:
reverseA_in_gga(
x1,
x2,
x3) =
reverseA_in_gga(
x1,
x2)
[] =
[]
reverseA_out_gga(
x1,
x2,
x3) =
reverseA_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x3,
x4,
x6)
(3) 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:
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42, T45) → U1_GGA(T43, T40, T41, T42, T45, reverseA_in_gga(T41, T42, .(T40, .(T43, T45))))
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42, T45) → REVERSEA_IN_GGA(T41, T42, .(T40, .(T43, T45)))
The TRS R consists of the following rules:
reverseA_in_gga([], T5, T5) → reverseA_out_gga([], T5, T5)
reverseA_in_gga(.(T28, []), .(T28, T29), T29) → reverseA_out_gga(.(T28, []), .(T28, T29), T29)
reverseA_in_gga(.(T43, .(T40, T41)), T42, T45) → U1_gga(T43, T40, T41, T42, T45, reverseA_in_gga(T41, T42, .(T40, .(T43, T45))))
U1_gga(T43, T40, T41, T42, T45, reverseA_out_gga(T41, T42, .(T40, .(T43, T45)))) → reverseA_out_gga(.(T43, .(T40, T41)), T42, T45)
The argument filtering Pi contains the following mapping:
reverseA_in_gga(
x1,
x2,
x3) =
reverseA_in_gga(
x1,
x2)
[] =
[]
reverseA_out_gga(
x1,
x2,
x3) =
reverseA_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x3,
x4,
x6)
REVERSEA_IN_GGA(
x1,
x2,
x3) =
REVERSEA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GGA(
x1,
x2,
x3,
x4,
x6)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42, T45) → U1_GGA(T43, T40, T41, T42, T45, reverseA_in_gga(T41, T42, .(T40, .(T43, T45))))
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42, T45) → REVERSEA_IN_GGA(T41, T42, .(T40, .(T43, T45)))
The TRS R consists of the following rules:
reverseA_in_gga([], T5, T5) → reverseA_out_gga([], T5, T5)
reverseA_in_gga(.(T28, []), .(T28, T29), T29) → reverseA_out_gga(.(T28, []), .(T28, T29), T29)
reverseA_in_gga(.(T43, .(T40, T41)), T42, T45) → U1_gga(T43, T40, T41, T42, T45, reverseA_in_gga(T41, T42, .(T40, .(T43, T45))))
U1_gga(T43, T40, T41, T42, T45, reverseA_out_gga(T41, T42, .(T40, .(T43, T45)))) → reverseA_out_gga(.(T43, .(T40, T41)), T42, T45)
The argument filtering Pi contains the following mapping:
reverseA_in_gga(
x1,
x2,
x3) =
reverseA_in_gga(
x1,
x2)
[] =
[]
reverseA_out_gga(
x1,
x2,
x3) =
reverseA_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x3,
x4,
x6)
REVERSEA_IN_GGA(
x1,
x2,
x3) =
REVERSEA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GGA(
x1,
x2,
x3,
x4,
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:
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42, T45) → REVERSEA_IN_GGA(T41, T42, .(T40, .(T43, T45)))
The TRS R consists of the following rules:
reverseA_in_gga([], T5, T5) → reverseA_out_gga([], T5, T5)
reverseA_in_gga(.(T28, []), .(T28, T29), T29) → reverseA_out_gga(.(T28, []), .(T28, T29), T29)
reverseA_in_gga(.(T43, .(T40, T41)), T42, T45) → U1_gga(T43, T40, T41, T42, T45, reverseA_in_gga(T41, T42, .(T40, .(T43, T45))))
U1_gga(T43, T40, T41, T42, T45, reverseA_out_gga(T41, T42, .(T40, .(T43, T45)))) → reverseA_out_gga(.(T43, .(T40, T41)), T42, T45)
The argument filtering Pi contains the following mapping:
reverseA_in_gga(
x1,
x2,
x3) =
reverseA_in_gga(
x1,
x2)
[] =
[]
reverseA_out_gga(
x1,
x2,
x3) =
reverseA_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gga(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_gga(
x1,
x2,
x3,
x4,
x6)
REVERSEA_IN_GGA(
x1,
x2,
x3) =
REVERSEA_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(7) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42, T45) → REVERSEA_IN_GGA(T41, T42, .(T40, .(T43, T45)))
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
REVERSEA_IN_GGA(
x1,
x2,
x3) =
REVERSEA_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(9) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42) → REVERSEA_IN_GGA(T41, T42)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) 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:
- REVERSEA_IN_GGA(.(T43, .(T40, T41)), T42) → REVERSEA_IN_GGA(T41, T42)
The graph contains the following edges 1 > 1, 2 >= 2
(12) YES