Left Termination of the query pattern
sublist_in_2(g, g)
w.r.t. the given Prolog program could not be shown:
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
Clauses:
append1([], Ys, Ys).
append1(.(X, Xs), Ys, .(X, Zs)) :- append1(Xs, Ys, Zs).
append2([], Ys, Ys).
append2(.(X, Xs), Ys, .(X, Zs)) :- append2(Xs, Ys, Zs).
sublist(X, Y) :- ','(append1(U, X, V), append2(V, W, Y)).
Queries:
sublist(g,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:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x3)
U4(x1, x2, x3) = U4(x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
append2_out(x1, x2, x3) = append2_out(x2)
sublist_out(x1, x2) = sublist_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:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x3)
U4(x1, x2, x3) = U4(x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
append2_out(x1, x2, x3) = append2_out(x2)
sublist_out(x1, x2) = sublist_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:
SUBLIST_IN(X, Y) → U31(X, Y, append1_in(U, X, V))
SUBLIST_IN(X, Y) → APPEND1_IN(U, X, V)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
U31(X, Y, append1_out(U, X, V)) → U41(X, Y, append2_in(V, W, Y))
U31(X, Y, append1_out(U, X, V)) → APPEND2_IN(V, W, Y)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x3)
U4(x1, x2, x3) = U4(x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
append2_out(x1, x2, x3) = append2_out(x2)
sublist_out(x1, x2) = sublist_out
APPEND1_IN(x1, x2, x3) = APPEND1_IN(x2)
U41(x1, x2, x3) = U41(x3)
U31(x1, x2, x3) = U31(x2, x3)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
U21(x1, x2, x3, x4, x5) = U21(x5)
U11(x1, x2, x3, x4, x5) = U11(x5)
SUBLIST_IN(x1, x2) = SUBLIST_IN(x1, x2)
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:
SUBLIST_IN(X, Y) → U31(X, Y, append1_in(U, X, V))
SUBLIST_IN(X, Y) → APPEND1_IN(U, X, V)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
U31(X, Y, append1_out(U, X, V)) → U41(X, Y, append2_in(V, W, Y))
U31(X, Y, append1_out(U, X, V)) → APPEND2_IN(V, W, Y)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x3)
U4(x1, x2, x3) = U4(x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
append2_out(x1, x2, x3) = append2_out(x2)
sublist_out(x1, x2) = sublist_out
APPEND1_IN(x1, x2, x3) = APPEND1_IN(x2)
U41(x1, x2, x3) = U41(x3)
U31(x1, x2, x3) = U31(x2, x3)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
U21(x1, x2, x3, x4, x5) = U21(x5)
U11(x1, x2, x3, x4, x5) = U11(x5)
SUBLIST_IN(x1, x2) = SUBLIST_IN(x1, x2)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x3)
U4(x1, x2, x3) = U4(x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
append2_out(x1, x2, x3) = append2_out(x2)
sublist_out(x1, x2) = sublist_out
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
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:
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
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
↳ QDPSizeChangeProof
↳ PiDP
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
APPEND2_IN(.(Xs), .(Zs)) → APPEND2_IN(Xs, Zs)
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:
- APPEND2_IN(.(Xs), .(Zs)) → APPEND2_IN(Xs, Zs)
The graph contains the following edges 1 > 1, 2 > 2
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x3)
U4(x1, x2, x3) = U4(x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
append2_out(x1, x2, x3) = append2_out(x2)
sublist_out(x1, x2) = sublist_out
APPEND1_IN(x1, x2, x3) = APPEND1_IN(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
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(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
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ NonTerminationProof
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
APPEND1_IN(Ys) → APPEND1_IN(Ys)
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:
APPEND1_IN(Ys) → APPEND1_IN(Ys)
The TRS R consists of the following rules:none
s = APPEND1_IN(Ys) evaluates to t =APPEND1_IN(Ys)
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 APPEND1_IN(Ys) to APPEND1_IN(Ys).
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:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x1, x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x2, x3)
U4(x1, x2, x3) = U4(x1, x2, x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x2, x4, x5)
append2_out(x1, x2, x3) = append2_out(x1, x2, x3)
sublist_out(x1, x2) = sublist_out(x1, x2)
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:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x1, x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x2, x3)
U4(x1, x2, x3) = U4(x1, x2, x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x2, x4, x5)
append2_out(x1, x2, x3) = append2_out(x1, x2, x3)
sublist_out(x1, x2) = sublist_out(x1, x2)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
SUBLIST_IN(X, Y) → U31(X, Y, append1_in(U, X, V))
SUBLIST_IN(X, Y) → APPEND1_IN(U, X, V)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
U31(X, Y, append1_out(U, X, V)) → U41(X, Y, append2_in(V, W, Y))
U31(X, Y, append1_out(U, X, V)) → APPEND2_IN(V, W, Y)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x1, x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x2, x3)
U4(x1, x2, x3) = U4(x1, x2, x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x2, x4, x5)
append2_out(x1, x2, x3) = append2_out(x1, x2, x3)
sublist_out(x1, x2) = sublist_out(x1, x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(x2)
U41(x1, x2, x3) = U41(x1, x2, x3)
U31(x1, x2, x3) = U31(x1, x2, x3)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
U21(x1, x2, x3, x4, x5) = U21(x2, x4, x5)
U11(x1, x2, x3, x4, x5) = U11(x3, x5)
SUBLIST_IN(x1, x2) = SUBLIST_IN(x1, x2)
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:
SUBLIST_IN(X, Y) → U31(X, Y, append1_in(U, X, V))
SUBLIST_IN(X, Y) → APPEND1_IN(U, X, V)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
U31(X, Y, append1_out(U, X, V)) → U41(X, Y, append2_in(V, W, Y))
U31(X, Y, append1_out(U, X, V)) → APPEND2_IN(V, W, Y)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x1, x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x2, x3)
U4(x1, x2, x3) = U4(x1, x2, x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x2, x4, x5)
append2_out(x1, x2, x3) = append2_out(x1, x2, x3)
sublist_out(x1, x2) = sublist_out(x1, x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(x2)
U41(x1, x2, x3) = U41(x1, x2, x3)
U31(x1, x2, x3) = U31(x1, x2, x3)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
U21(x1, x2, x3, x4, x5) = U21(x2, x4, x5)
U11(x1, x2, x3, x4, x5) = U11(x3, x5)
SUBLIST_IN(x1, x2) = SUBLIST_IN(x1, x2)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ UsableRulesProof
↳ PiDP
Pi DP problem:
The TRS P consists of the following rules:
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x1, x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x2, x3)
U4(x1, x2, x3) = U4(x1, x2, x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x2, x4, x5)
append2_out(x1, x2, x3) = append2_out(x1, x2, x3)
sublist_out(x1, x2) = sublist_out(x1, x2)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
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:
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APPEND2_IN(x1, x2, x3) = APPEND2_IN(x1, x3)
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
↳ QDPSizeChangeProof
↳ PiDP
Q DP problem:
The TRS P consists of the following rules:
APPEND2_IN(.(Xs), .(Zs)) → APPEND2_IN(Xs, Zs)
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:
- APPEND2_IN(.(Xs), .(Zs)) → APPEND2_IN(Xs, Zs)
The graph contains the following edges 1 > 1, 2 > 2
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
The TRS R consists of the following rules:
sublist_in(X, Y) → U3(X, Y, append1_in(U, X, V))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(U, X, V)) → U4(X, Y, append2_in(V, W, Y))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(V, W, Y)) → sublist_out(X, Y)
The argument filtering Pi contains the following mapping:
sublist_in(x1, x2) = sublist_in(x1, x2)
U3(x1, x2, x3) = U3(x1, x2, x3)
append1_in(x1, x2, x3) = append1_in(x2)
.(x1, x2) = .(x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
[] = []
append1_out(x1, x2, x3) = append1_out(x1, x2, x3)
U4(x1, x2, x3) = U4(x1, x2, x3)
append2_in(x1, x2, x3) = append2_in(x1, x3)
U2(x1, x2, x3, x4, x5) = U2(x2, x4, x5)
append2_out(x1, x2, x3) = append2_out(x1, x2, x3)
sublist_out(x1, x2) = sublist_out(x1, x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APPEND1_IN(x1, x2, x3) = APPEND1_IN(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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ AND
↳ PiDP
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
APPEND1_IN(Ys) → APPEND1_IN(Ys)
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:
APPEND1_IN(Ys) → APPEND1_IN(Ys)
The TRS R consists of the following rules:none
s = APPEND1_IN(Ys) evaluates to t =APPEND1_IN(Ys)
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 APPEND1_IN(Ys) to APPEND1_IN(Ys).