(0) Obligation:
Clauses:len([], 0).
len(.(X1, Ts), s(N)) :- len(Ts, N).
Queries:
len(g,a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:len1([], 0).
len1(.(T6, []), s(0)).
len1(.(T6, .(T16, T17)), s(s(T19))) :- len1(T17, T19).
Queries:
len1(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: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(.(T6, []), s(0)) → len1_out_ga(.(T6, []), s(0))
len1_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, len1_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, len1_out_ga(T17, T19)) → len1_out_ga(.(T6, .(T16, T17)), s(s(T19)))
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, x5) = U1_ga(x5)
s(x1) = s(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T6, []), s(0)) → len1_out_ga(.(T6, []), s(0))
len1_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, len1_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, len1_out_ga(T17, T19)) → len1_out_ga(.(T6, .(T16, T17)), s(s(T19)))
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, x5) = U1_ga(x5)
s(x1) = s(x1)
(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:
LEN1_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → U1_GA(T6, T16, T17, T19, len1_in_ga(T17, T19))
LEN1_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → LEN1_IN_GA(T17, T19)
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T6, []), s(0)) → len1_out_ga(.(T6, []), s(0))
len1_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, len1_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, len1_out_ga(T17, T19)) → len1_out_ga(.(T6, .(T16, T17)), s(s(T19)))
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, x5) = U1_ga(x5)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5) = U1_GA(x5)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
LEN1_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → U1_GA(T6, T16, T17, T19, len1_in_ga(T17, T19))
LEN1_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → LEN1_IN_GA(T17, T19)
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T6, []), s(0)) → len1_out_ga(.(T6, []), s(0))
len1_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, len1_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, len1_out_ga(T17, T19)) → len1_out_ga(.(T6, .(T16, T17)), s(s(T19)))
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, x5) = U1_ga(x5)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5) = U1_GA(x5)
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:
LEN1_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → LEN1_IN_GA(T17, T19)
The TRS R consists of the following rules:
len1_in_ga([], 0) → len1_out_ga([], 0)
len1_in_ga(.(T6, []), s(0)) → len1_out_ga(.(T6, []), s(0))
len1_in_ga(.(T6, .(T16, T17)), s(s(T19))) → U1_ga(T6, T16, T17, T19, len1_in_ga(T17, T19))
U1_ga(T6, T16, T17, T19, len1_out_ga(T17, T19)) → len1_out_ga(.(T6, .(T16, T17)), s(s(T19)))
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, x5) = U1_ga(x5)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(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:
LEN1_IN_GA(.(T6, .(T16, T17)), s(s(T19))) → LEN1_IN_GA(T17, T19)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
LEN1_IN_GA(x1, x2) = LEN1_IN_GA(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:
LEN1_IN_GA(.(T6, .(T16, T17))) → LEN1_IN_GA(T17)
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:
- LEN1_IN_GA(.(T6, .(T16, T17))) → LEN1_IN_GA(T17)
The graph contains the following edges 1 > 1