Left Termination of the query pattern
prefix_in_2(g, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
prefix(Xs, Ys) :- app(Xs, X, Ys).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
Queries:
prefix(g,a).
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
prefix_in: (b,f)
app_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
prefix_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_gaa(Xs, X, Ys))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U2_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U2_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_gaa(Xs, X, Ys)) → prefix_out_ga(Xs, Ys)
The argument filtering Pi contains the following mapping:
prefix_in_ga(x1, x2) = prefix_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
app_in_gaa(x1, x2, x3) = app_in_gaa(x1)
[] = []
app_out_gaa(x1, x2, x3) = app_out_gaa
.(x1, x2) = .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
prefix_out_ga(x1, x2) = prefix_out_ga
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:
prefix_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_gaa(Xs, X, Ys))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U2_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U2_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_gaa(Xs, X, Ys)) → prefix_out_ga(Xs, Ys)
The argument filtering Pi contains the following mapping:
prefix_in_ga(x1, x2) = prefix_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
app_in_gaa(x1, x2, x3) = app_in_gaa(x1)
[] = []
app_out_gaa(x1, x2, x3) = app_out_gaa
.(x1, x2) = .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
prefix_out_ga(x1, x2) = prefix_out_ga
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
PREFIX_IN_GA(Xs, Ys) → U1_GA(Xs, Ys, app_in_gaa(Xs, X, Ys))
PREFIX_IN_GA(Xs, Ys) → APP_IN_GAA(Xs, X, Ys)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U2_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)
The TRS R consists of the following rules:
prefix_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_gaa(Xs, X, Ys))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U2_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U2_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_gaa(Xs, X, Ys)) → prefix_out_ga(Xs, Ys)
The argument filtering Pi contains the following mapping:
prefix_in_ga(x1, x2) = prefix_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
app_in_gaa(x1, x2, x3) = app_in_gaa(x1)
[] = []
app_out_gaa(x1, x2, x3) = app_out_gaa
.(x1, x2) = .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
prefix_out_ga(x1, x2) = prefix_out_ga
U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5)
PREFIX_IN_GA(x1, x2) = PREFIX_IN_GA(x1)
U1_GA(x1, x2, x3) = U1_GA(x3)
APP_IN_GAA(x1, x2, x3) = APP_IN_GAA(x1)
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:
PREFIX_IN_GA(Xs, Ys) → U1_GA(Xs, Ys, app_in_gaa(Xs, X, Ys))
PREFIX_IN_GA(Xs, Ys) → APP_IN_GAA(Xs, X, Ys)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U2_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)
The TRS R consists of the following rules:
prefix_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_gaa(Xs, X, Ys))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U2_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U2_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_gaa(Xs, X, Ys)) → prefix_out_ga(Xs, Ys)
The argument filtering Pi contains the following mapping:
prefix_in_ga(x1, x2) = prefix_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
app_in_gaa(x1, x2, x3) = app_in_gaa(x1)
[] = []
app_out_gaa(x1, x2, x3) = app_out_gaa
.(x1, x2) = .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
prefix_out_ga(x1, x2) = prefix_out_ga
U2_GAA(x1, x2, x3, x4, x5) = U2_GAA(x5)
PREFIX_IN_GA(x1, x2) = PREFIX_IN_GA(x1)
U1_GA(x1, x2, x3) = U1_GA(x3)
APP_IN_GAA(x1, x2, x3) = APP_IN_GAA(x1)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)
The TRS R consists of the following rules:
prefix_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_gaa(Xs, X, Ys))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U2_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U2_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_gaa(Xs, X, Ys)) → prefix_out_ga(Xs, Ys)
The argument filtering Pi contains the following mapping:
prefix_in_ga(x1, x2) = prefix_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
app_in_gaa(x1, x2, x3) = app_in_gaa(x1)
[] = []
app_out_gaa(x1, x2, x3) = app_out_gaa
.(x1, x2) = .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5) = U2_gaa(x5)
prefix_out_ga(x1, x2) = prefix_out_ga
APP_IN_GAA(x1, x2, x3) = APP_IN_GAA(x1)
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:
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APP_IN_GAA(x1, x2, x3) = APP_IN_GAA(x1)
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:
APP_IN_GAA(.(X, Xs)) → APP_IN_GAA(Xs)
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:
- APP_IN_GAA(.(X, Xs)) → APP_IN_GAA(Xs)
The graph contains the following edges 1 > 1