(0) Obligation:
Clauses:
reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).
Queries:
reverse(g,a).
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (b,f)
reverse_in: (b,b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x2)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x2)
(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:
REVERSE_IN_GA(X1s, X2s) → U1_GA(X1s, X2s, reverse_in_gga(X1s, [], X2s))
REVERSE_IN_GA(X1s, X2s) → REVERSE_IN_GGA(X1s, [], X2s)
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → U2_GGA(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x2)
REVERSE_IN_GA(
x1,
x2) =
REVERSE_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x3)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_IN_GGA(
x1,
x2)
U2_GGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGA(
x5)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN_GA(X1s, X2s) → U1_GA(X1s, X2s, reverse_in_gga(X1s, [], X2s))
REVERSE_IN_GA(X1s, X2s) → REVERSE_IN_GGA(X1s, [], X2s)
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → U2_GGA(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x2)
REVERSE_IN_GA(
x1,
x2) =
REVERSE_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x3)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_IN_GGA(
x1,
x2)
U2_GGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGA(
x5)
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:
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x2)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_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:
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_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:
REVERSE_IN_GGA(.(X, X1s), X2s) → REVERSE_IN_GGA(X1s, .(X, X2s))
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:
- REVERSE_IN_GGA(.(X, X1s), X2s) → REVERSE_IN_GGA(X1s, .(X, X2s))
The graph contains the following edges 1 > 1
(12) TRUE
(13) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (b,f)
reverse_in: (b,b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x1,
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x1,
x2,
x3,
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x1,
x2)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(14) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x1,
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x1,
x2,
x3,
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x1,
x2)
(15) 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:
REVERSE_IN_GA(X1s, X2s) → U1_GA(X1s, X2s, reverse_in_gga(X1s, [], X2s))
REVERSE_IN_GA(X1s, X2s) → REVERSE_IN_GGA(X1s, [], X2s)
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → U2_GGA(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x1,
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x1,
x2,
x3,
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x1,
x2)
REVERSE_IN_GA(
x1,
x2) =
REVERSE_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x1,
x3)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_IN_GGA(
x1,
x2)
U2_GGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGA(
x1,
x2,
x3,
x5)
We have to consider all (P,R,Pi)-chains
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN_GA(X1s, X2s) → U1_GA(X1s, X2s, reverse_in_gga(X1s, [], X2s))
REVERSE_IN_GA(X1s, X2s) → REVERSE_IN_GGA(X1s, [], X2s)
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → U2_GGA(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x1,
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x1,
x2,
x3,
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x1,
x2)
REVERSE_IN_GA(
x1,
x2) =
REVERSE_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x1,
x3)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_IN_GGA(
x1,
x2)
U2_GGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGA(
x1,
x2,
x3,
x5)
We have to consider all (P,R,Pi)-chains
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
(18) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
reverse_in_ga(X1s, X2s) → U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s))
reverse_in_gga([], Xs, Xs) → reverse_out_gga([], Xs, Xs)
reverse_in_gga(.(X, X1s), X2s, Ys) → U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys))
U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) → reverse_out_gga(.(X, X1s), X2s, Ys)
U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) → reverse_out_ga(X1s, X2s)
The argument filtering Pi contains the following mapping:
reverse_in_ga(
x1,
x2) =
reverse_in_ga(
x1)
U1_ga(
x1,
x2,
x3) =
U1_ga(
x1,
x3)
reverse_in_gga(
x1,
x2,
x3) =
reverse_in_gga(
x1,
x2)
[] =
[]
reverse_out_gga(
x1,
x2,
x3) =
reverse_out_gga(
x1,
x2,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U2_gga(
x1,
x2,
x3,
x4,
x5) =
U2_gga(
x1,
x2,
x3,
x5)
reverse_out_ga(
x1,
x2) =
reverse_out_ga(
x1,
x2)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(19) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(20) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE_IN_GGA(.(X, X1s), X2s, Ys) → REVERSE_IN_GGA(X1s, .(X, X2s), Ys)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
REVERSE_IN_GGA(
x1,
x2,
x3) =
REVERSE_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(21) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REVERSE_IN_GGA(.(X, X1s), X2s) → REVERSE_IN_GGA(X1s, .(X, X2s))
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.