Left Termination of the query pattern
fl_in_3(a, a, g)
w.r.t. the given Prolog program could not be shown:
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
Clauses:
fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).
Queries:
fl(a,a,g).
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:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PrologToPiTRSProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out
U31(x1, x2, x3, x4, x5) = U31(x5)
APPEND_IN(x1, x2, x3) = APPEND_IN
FL_IN(x1, x2, x3) = FL_IN(x3)
U21(x1, x2, x3, x4, x5) = U21(x5)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out
U31(x1, x2, x3, x4, x5) = U31(x5)
APPEND_IN(x1, x2, x3) = APPEND_IN
FL_IN(x1, x2, x3) = FL_IN(x3)
U21(x1, x2, x3, x4, x5) = U21(x5)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out
APPEND_IN(x1, x2, x3) = APPEND_IN
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
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PiDP
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APPEND_IN(x1, x2, x3) = APPEND_IN
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
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ NonTerminationProof
↳ PiDP
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
APPEND_IN → APPEND_IN
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
The TRS P consists of the following rules:
APPEND_IN → APPEND_IN
The TRS R consists of the following rules:none
s = APPEND_IN evaluates to t =APPEND_IN
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from APPEND_IN to APPEND_IN.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x5)
0 = 0
fl_out(x1, x2, x3) = fl_out
FL_IN(x1, x2, x3) = FL_IN(x3)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
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
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
FL_IN(x1, x2, x3) = FL_IN(x3)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
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
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ QDPSizeChangeProof
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
U11(Z, append_out) → FL_IN(Z)
FL_IN(s(Z)) → U11(Z, append_in)
The TRS R consists of the following rules:
append_in → U3(append_in)
append_in → append_out
U3(append_out) → append_out
The set Q consists of the following terms:
append_in
U3(x0)
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:
- U11(Z, append_out) → FL_IN(Z)
The graph contains the following edges 1 >= 1
- FL_IN(s(Z)) → U11(Z, append_in)
The graph contains the following edges 1 > 1
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:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x4, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x4, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(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:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x4, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
U31(x1, x2, x3, x4, x5) = U31(x5)
APPEND_IN(x1, x2, x3) = APPEND_IN
FL_IN(x1, x2, x3) = FL_IN(x3)
U21(x1, x2, x3, x4, x5) = U21(x4, x5)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x4, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
U31(x1, x2, x3, x4, x5) = U31(x5)
APPEND_IN(x1, x2, x3) = APPEND_IN
FL_IN(x1, x2, x3) = FL_IN(x3)
U21(x1, x2, x3, x4, x5) = U21(x4, x5)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x4, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
APPEND_IN(x1, x2, x3) = APPEND_IN
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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PiDP
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
APPEND_IN(x1, x2, x3) = APPEND_IN
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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ NonTerminationProof
↳ PiDP
Q DP problem:
The TRS P consists of the following rules:
APPEND_IN → APPEND_IN
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
The TRS P consists of the following rules:
APPEND_IN → APPEND_IN
The TRS R consists of the following rules:none
s = APPEND_IN evaluates to t =APPEND_IN
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from APPEND_IN to APPEND_IN.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3) = fl_in(x3)
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
U1(x1, x2, x3, x4, x5) = U1(x4, x5)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
U2(x1, x2, x3, x4, x5) = U2(x4, x5)
0 = 0
fl_out(x1, x2, x3) = fl_out(x3)
FL_IN(x1, x2, x3) = FL_IN(x3)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)
The TRS R consists of the following rules:
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
s(x1) = s(x1)
append_in(x1, x2, x3) = append_in
U3(x1, x2, x3, x4, x5) = U3(x5)
[] = []
append_out(x1, x2, x3) = append_out
FL_IN(x1, x2, x3) = FL_IN(x3)
U11(x1, x2, x3, x4, x5) = U11(x4, x5)
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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
U11(Z, append_out) → FL_IN(Z)
FL_IN(s(Z)) → U11(Z, append_in)
The TRS R consists of the following rules:
append_in → U3(append_in)
append_in → append_out
U3(append_out) → append_out
The set Q consists of the following terms:
append_in
U3(x0)
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:
- U11(Z, append_out) → FL_IN(Z)
The graph contains the following edges 1 >= 1
- FL_IN(s(Z)) → U11(Z, append_in)
The graph contains the following edges 1 > 1