(0) Obligation:
Clauses:
len1([], 0).
len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M))).
eq(X, X).
Queries:
len1(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:
len1_in: (b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1)
s(
x1) =
s(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1)
s(
x1) =
s(
x1)
(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:
LEN1_IN_GA(.(X1, Ts), N) → U1_GA(X1, Ts, N, len1_in_ga(Ts, M))
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → U2_GA(X1, Ts, N, M, eq_in_ag(N, s(M)))
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → EQ_IN_AG(N, s(M))
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1)
s(
x1) =
s(
x1)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x4)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x5)
EQ_IN_AG(
x1,
x2) =
EQ_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LEN1_IN_GA(.(X1, Ts), N) → U1_GA(X1, Ts, N, len1_in_ga(Ts, M))
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → U2_GA(X1, Ts, N, M, eq_in_ag(N, s(M)))
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → EQ_IN_AG(N, s(M))
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1)
s(
x1) =
s(
x1)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x4)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x5)
EQ_IN_AG(
x1,
x2) =
EQ_IN_AG(
x2)
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:
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1)
s(
x1) =
s(
x1)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
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:
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
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:
LEN1_IN_GA(.(X1, Ts)) → LEN1_IN_GA(Ts)
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:
- LEN1_IN_GA(.(X1, Ts)) → LEN1_IN_GA(Ts)
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:
len1_in: (b,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x2,
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x1,
x2,
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1,
x2)
s(
x1) =
s(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(14) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x2,
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x1,
x2,
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1,
x2)
s(
x1) =
s(
x1)
(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:
LEN1_IN_GA(.(X1, Ts), N) → U1_GA(X1, Ts, N, len1_in_ga(Ts, M))
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → U2_GA(X1, Ts, N, M, eq_in_ag(N, s(M)))
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → EQ_IN_AG(N, s(M))
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x2,
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x1,
x2,
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1,
x2)
s(
x1) =
s(
x1)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x1,
x2,
x4)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x5)
EQ_IN_AG(
x1,
x2) =
EQ_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LEN1_IN_GA(.(X1, Ts), N) → U1_GA(X1, Ts, N, len1_in_ga(Ts, M))
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → U2_GA(X1, Ts, N, M, eq_in_ag(N, s(M)))
U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) → EQ_IN_AG(N, s(M))
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x2,
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x1,
x2,
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1,
x2)
s(
x1) =
s(
x1)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
U1_GA(
x1,
x2,
x3,
x4) =
U1_GA(
x1,
x2,
x4)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x5)
EQ_IN_AG(
x1,
x2) =
EQ_IN_AG(
x2)
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:
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(X1, Ts), N) → U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) → U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
eq_in_ag(X, X) → eq_out_ag(X, X)
U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) → len1_out_ga(.(X1, Ts), N)
The argument filtering Pi contains the following mapping:
len1_in_ga(
x1,
x2) =
len1_in_ga(
x1)
[] =
[]
len1_out_ga(
x1,
x2) =
len1_out_ga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
U1_ga(
x1,
x2,
x3,
x4) =
U1_ga(
x1,
x2,
x4)
U2_ga(
x1,
x2,
x3,
x4,
x5) =
U2_ga(
x1,
x2,
x5)
eq_in_ag(
x1,
x2) =
eq_in_ag(
x2)
eq_out_ag(
x1,
x2) =
eq_out_ag(
x1,
x2)
s(
x1) =
s(
x1)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
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:
LEN1_IN_GA(.(X1, Ts), N) → LEN1_IN_GA(Ts, M)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
LEN1_IN_GA(
x1,
x2) =
LEN1_IN_GA(
x1)
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:
LEN1_IN_GA(.(X1, Ts)) → LEN1_IN_GA(Ts)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.