Left Termination of the query pattern
append_in_3(g, g, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z).
append([], Y, Y).
Queries:
append(g,g,a).
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
append_in: (b,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
append_in_gga(.(H, X), Y, .(X, Z)) → U1_gga(H, X, Y, Z, append_in_gga(X, Y, Z))
append_in_gga([], Y, Y) → append_out_gga([], Y, Y)
U1_gga(H, X, Y, Z, append_out_gga(X, Y, Z)) → append_out_gga(.(H, X), Y, .(X, Z))
The argument filtering Pi contains the following mapping:
append_in_gga(x1, x2, x3) = append_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x2, x5)
[] = []
append_out_gga(x1, x2, x3) = append_out_gga(x3)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
append_in_gga(.(H, X), Y, .(X, Z)) → U1_gga(H, X, Y, Z, append_in_gga(X, Y, Z))
append_in_gga([], Y, Y) → append_out_gga([], Y, Y)
U1_gga(H, X, Y, Z, append_out_gga(X, Y, Z)) → append_out_gga(.(H, X), Y, .(X, Z))
The argument filtering Pi contains the following mapping:
append_in_gga(x1, x2, x3) = append_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x2, x5)
[] = []
append_out_gga(x1, x2, x3) = append_out_gga(x3)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GGA(.(H, X), Y, .(X, Z)) → U1_GGA(H, X, Y, Z, append_in_gga(X, Y, Z))
APPEND_IN_GGA(.(H, X), Y, .(X, Z)) → APPEND_IN_GGA(X, Y, Z)
The TRS R consists of the following rules:
append_in_gga(.(H, X), Y, .(X, Z)) → U1_gga(H, X, Y, Z, append_in_gga(X, Y, Z))
append_in_gga([], Y, Y) → append_out_gga([], Y, Y)
U1_gga(H, X, Y, Z, append_out_gga(X, Y, Z)) → append_out_gga(.(H, X), Y, .(X, Z))
The argument filtering Pi contains the following mapping:
append_in_gga(x1, x2, x3) = append_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x2, x5)
[] = []
append_out_gga(x1, x2, x3) = append_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x2, x5)
APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GGA(.(H, X), Y, .(X, Z)) → U1_GGA(H, X, Y, Z, append_in_gga(X, Y, Z))
APPEND_IN_GGA(.(H, X), Y, .(X, Z)) → APPEND_IN_GGA(X, Y, Z)
The TRS R consists of the following rules:
append_in_gga(.(H, X), Y, .(X, Z)) → U1_gga(H, X, Y, Z, append_in_gga(X, Y, Z))
append_in_gga([], Y, Y) → append_out_gga([], Y, Y)
U1_gga(H, X, Y, Z, append_out_gga(X, Y, Z)) → append_out_gga(.(H, X), Y, .(X, Z))
The argument filtering Pi contains the following mapping:
append_in_gga(x1, x2, x3) = append_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x2, x5)
[] = []
append_out_gga(x1, x2, x3) = append_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x2, x5)
APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GGA(.(H, X), Y, .(X, Z)) → APPEND_IN_GGA(X, Y, Z)
The TRS R consists of the following rules:
append_in_gga(.(H, X), Y, .(X, Z)) → U1_gga(H, X, Y, Z, append_in_gga(X, Y, Z))
append_in_gga([], Y, Y) → append_out_gga([], Y, Y)
U1_gga(H, X, Y, Z, append_out_gga(X, Y, Z)) → append_out_gga(.(H, X), Y, .(X, Z))
The argument filtering Pi contains the following mapping:
append_in_gga(x1, x2, x3) = append_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x2, x5)
[] = []
append_out_gga(x1, x2, x3) = append_out_gga(x3)
APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2)
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GGA(.(H, X), Y, .(X, Z)) → APPEND_IN_GGA(X, Y, Z)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APPEND_IN_GGA(x1, x2, x3) = APPEND_IN_GGA(x1, x2)
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
APPEND_IN_GGA(.(H, X), Y) → APPEND_IN_GGA(X, Y)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- APPEND_IN_GGA(.(H, X), Y) → APPEND_IN_GGA(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2