Left Termination of the query pattern
app3_b_in_4(g, g, g, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
app3_a(Xs, Ys, Zs, Us) :- ','(app(Xs, Ys, Vs), app(Vs, Zs, Us)).
app3_b(Xs, Ys, Zs, Us) :- ','(app(Ys, Zs, Vs), app(Xs, Vs, Us)).
app([], Ys, Ys).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
Queries:
app3_b(g,g,g,a).
We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
app3_b_in(Xs, Ys, Zs, Us) → U3(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U5(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
U5(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U4(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U4(Xs, Ys, Zs, Us, app_out(Xs, Vs, Us)) → app3_b_out(Xs, Ys, Zs, Us)
The argument filtering Pi contains the following mapping:
app3_b_in(x1, x2, x3, x4) = app3_b_in(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
app_in(x1, x2, x3) = app_in(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
[] = []
app_out(x1, x2, x3) = app_out(x3)
U4(x1, x2, x3, x4, x5) = U4(x5)
app3_b_out(x1, x2, x3, x4) = app3_b_out(x4)
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:
app3_b_in(Xs, Ys, Zs, Us) → U3(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U5(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
U5(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U4(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U4(Xs, Ys, Zs, Us, app_out(Xs, Vs, Us)) → app3_b_out(Xs, Ys, Zs, Us)
The argument filtering Pi contains the following mapping:
app3_b_in(x1, x2, x3, x4) = app3_b_in(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
app_in(x1, x2, x3) = app_in(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
[] = []
app_out(x1, x2, x3) = app_out(x3)
U4(x1, x2, x3, x4, x5) = U4(x5)
app3_b_out(x1, x2, x3, x4) = app3_b_out(x4)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
APP3_B_IN(Xs, Ys, Zs, Us) → U31(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
APP3_B_IN(Xs, Ys, Zs, Us) → APP_IN(Ys, Zs, Vs)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U51(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)
U31(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U41(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U31(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → APP_IN(Xs, Vs, Us)
The TRS R consists of the following rules:
app3_b_in(Xs, Ys, Zs, Us) → U3(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U5(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
U5(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U4(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U4(Xs, Ys, Zs, Us, app_out(Xs, Vs, Us)) → app3_b_out(Xs, Ys, Zs, Us)
The argument filtering Pi contains the following mapping:
app3_b_in(x1, x2, x3, x4) = app3_b_in(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
app_in(x1, x2, x3) = app_in(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
[] = []
app_out(x1, x2, x3) = app_out(x3)
U4(x1, x2, x3, x4, x5) = U4(x5)
app3_b_out(x1, x2, x3, x4) = app3_b_out(x4)
APP3_B_IN(x1, x2, x3, x4) = APP3_B_IN(x1, x2, x3)
U51(x1, x2, x3, x4, x5) = U51(x1, x5)
U41(x1, x2, x3, x4, x5) = U41(x5)
U31(x1, x2, x3, x4, x5) = U31(x1, x5)
APP_IN(x1, x2, x3) = APP_IN(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:
APP3_B_IN(Xs, Ys, Zs, Us) → U31(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
APP3_B_IN(Xs, Ys, Zs, Us) → APP_IN(Ys, Zs, Vs)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U51(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)
U31(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U41(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U31(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → APP_IN(Xs, Vs, Us)
The TRS R consists of the following rules:
app3_b_in(Xs, Ys, Zs, Us) → U3(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U5(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
U5(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U4(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U4(Xs, Ys, Zs, Us, app_out(Xs, Vs, Us)) → app3_b_out(Xs, Ys, Zs, Us)
The argument filtering Pi contains the following mapping:
app3_b_in(x1, x2, x3, x4) = app3_b_in(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
app_in(x1, x2, x3) = app_in(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
[] = []
app_out(x1, x2, x3) = app_out(x3)
U4(x1, x2, x3, x4, x5) = U4(x5)
app3_b_out(x1, x2, x3, x4) = app3_b_out(x4)
APP3_B_IN(x1, x2, x3, x4) = APP3_B_IN(x1, x2, x3)
U51(x1, x2, x3, x4, x5) = U51(x1, x5)
U41(x1, x2, x3, x4, x5) = U41(x5)
U31(x1, x2, x3, x4, x5) = U31(x1, x5)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
app3_b_in(Xs, Ys, Zs, Us) → U3(Xs, Ys, Zs, Us, app_in(Ys, Zs, Vs))
app_in(.(X, Xs), Ys, .(X, Zs)) → U5(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
U5(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, Ys, Zs, Us, app_out(Ys, Zs, Vs)) → U4(Xs, Ys, Zs, Us, app_in(Xs, Vs, Us))
U4(Xs, Ys, Zs, Us, app_out(Xs, Vs, Us)) → app3_b_out(Xs, Ys, Zs, Us)
The argument filtering Pi contains the following mapping:
app3_b_in(x1, x2, x3, x4) = app3_b_in(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x1, x5)
app_in(x1, x2, x3) = app_in(x1, x2)
.(x1, x2) = .(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
[] = []
app_out(x1, x2, x3) = app_out(x3)
U4(x1, x2, x3, x4, x5) = U4(x5)
app3_b_out(x1, x2, x3, x4) = app3_b_out(x4)
APP_IN(x1, x2, x3) = APP_IN(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:
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APP_IN(x1, x2, x3) = APP_IN(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:
APP_IN(.(X, Xs), Ys) → APP_IN(Xs, Ys)
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(.(X, Xs), Ys) → APP_IN(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2