Left Termination of the query pattern
fold_in_3(g, g, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
fold(X, .(Y, Ys), Z) :- ','(myop(X, Y, V), fold(V, Ys, Z)).
fold(X, [], X).
myop(a, b, c).
Queries:
fold(g,g,a).
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
fold_in: (b,b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
fold_in_gga(X, .(Y, Ys), Z) → U1_gga(X, Y, Ys, Z, myop_in_gga(X, Y, V))
myop_in_gga(a, b, c) → myop_out_gga(a, b, c)
U1_gga(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_gga(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
fold_in_gga(X, [], X) → fold_out_gga(X, [], X)
U2_gga(X, Y, Ys, Z, fold_out_gga(V, Ys, Z)) → fold_out_gga(X, .(Y, Ys), Z)
The argument filtering Pi contains the following mapping:
fold_in_gga(x1, x2, x3) = fold_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x3, x5)
myop_in_gga(x1, x2, x3) = myop_in_gga(x1, x2)
a = a
b = b
myop_out_gga(x1, x2, x3) = myop_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5)
[] = []
fold_out_gga(x1, x2, x3) = fold_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:
fold_in_gga(X, .(Y, Ys), Z) → U1_gga(X, Y, Ys, Z, myop_in_gga(X, Y, V))
myop_in_gga(a, b, c) → myop_out_gga(a, b, c)
U1_gga(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_gga(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
fold_in_gga(X, [], X) → fold_out_gga(X, [], X)
U2_gga(X, Y, Ys, Z, fold_out_gga(V, Ys, Z)) → fold_out_gga(X, .(Y, Ys), Z)
The argument filtering Pi contains the following mapping:
fold_in_gga(x1, x2, x3) = fold_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x3, x5)
myop_in_gga(x1, x2, x3) = myop_in_gga(x1, x2)
a = a
b = b
myop_out_gga(x1, x2, x3) = myop_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5)
[] = []
fold_out_gga(x1, x2, x3) = fold_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:
FOLD_IN_GGA(X, .(Y, Ys), Z) → U1_GGA(X, Y, Ys, Z, myop_in_gga(X, Y, V))
FOLD_IN_GGA(X, .(Y, Ys), Z) → MYOP_IN_GGA(X, Y, V)
U1_GGA(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_GGA(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
U1_GGA(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → FOLD_IN_GGA(V, Ys, Z)
The TRS R consists of the following rules:
fold_in_gga(X, .(Y, Ys), Z) → U1_gga(X, Y, Ys, Z, myop_in_gga(X, Y, V))
myop_in_gga(a, b, c) → myop_out_gga(a, b, c)
U1_gga(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_gga(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
fold_in_gga(X, [], X) → fold_out_gga(X, [], X)
U2_gga(X, Y, Ys, Z, fold_out_gga(V, Ys, Z)) → fold_out_gga(X, .(Y, Ys), Z)
The argument filtering Pi contains the following mapping:
fold_in_gga(x1, x2, x3) = fold_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x3, x5)
myop_in_gga(x1, x2, x3) = myop_in_gga(x1, x2)
a = a
b = b
myop_out_gga(x1, x2, x3) = myop_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5)
[] = []
fold_out_gga(x1, x2, x3) = fold_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x3, x5)
U2_GGA(x1, x2, x3, x4, x5) = U2_GGA(x5)
FOLD_IN_GGA(x1, x2, x3) = FOLD_IN_GGA(x1, x2)
MYOP_IN_GGA(x1, x2, x3) = MYOP_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:
FOLD_IN_GGA(X, .(Y, Ys), Z) → U1_GGA(X, Y, Ys, Z, myop_in_gga(X, Y, V))
FOLD_IN_GGA(X, .(Y, Ys), Z) → MYOP_IN_GGA(X, Y, V)
U1_GGA(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_GGA(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
U1_GGA(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → FOLD_IN_GGA(V, Ys, Z)
The TRS R consists of the following rules:
fold_in_gga(X, .(Y, Ys), Z) → U1_gga(X, Y, Ys, Z, myop_in_gga(X, Y, V))
myop_in_gga(a, b, c) → myop_out_gga(a, b, c)
U1_gga(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_gga(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
fold_in_gga(X, [], X) → fold_out_gga(X, [], X)
U2_gga(X, Y, Ys, Z, fold_out_gga(V, Ys, Z)) → fold_out_gga(X, .(Y, Ys), Z)
The argument filtering Pi contains the following mapping:
fold_in_gga(x1, x2, x3) = fold_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x3, x5)
myop_in_gga(x1, x2, x3) = myop_in_gga(x1, x2)
a = a
b = b
myop_out_gga(x1, x2, x3) = myop_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5)
[] = []
fold_out_gga(x1, x2, x3) = fold_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x3, x5)
U2_GGA(x1, x2, x3, x4, x5) = U2_GGA(x5)
FOLD_IN_GGA(x1, x2, x3) = FOLD_IN_GGA(x1, x2)
MYOP_IN_GGA(x1, x2, x3) = MYOP_IN_GGA(x1, x2)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
U1_GGA(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → FOLD_IN_GGA(V, Ys, Z)
FOLD_IN_GGA(X, .(Y, Ys), Z) → U1_GGA(X, Y, Ys, Z, myop_in_gga(X, Y, V))
The TRS R consists of the following rules:
fold_in_gga(X, .(Y, Ys), Z) → U1_gga(X, Y, Ys, Z, myop_in_gga(X, Y, V))
myop_in_gga(a, b, c) → myop_out_gga(a, b, c)
U1_gga(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → U2_gga(X, Y, Ys, Z, fold_in_gga(V, Ys, Z))
fold_in_gga(X, [], X) → fold_out_gga(X, [], X)
U2_gga(X, Y, Ys, Z, fold_out_gga(V, Ys, Z)) → fold_out_gga(X, .(Y, Ys), Z)
The argument filtering Pi contains the following mapping:
fold_in_gga(x1, x2, x3) = fold_in_gga(x1, x2)
.(x1, x2) = .(x1, x2)
U1_gga(x1, x2, x3, x4, x5) = U1_gga(x3, x5)
myop_in_gga(x1, x2, x3) = myop_in_gga(x1, x2)
a = a
b = b
myop_out_gga(x1, x2, x3) = myop_out_gga(x3)
U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5)
[] = []
fold_out_gga(x1, x2, x3) = fold_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x3, x5)
FOLD_IN_GGA(x1, x2, x3) = FOLD_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:
U1_GGA(X, Y, Ys, Z, myop_out_gga(X, Y, V)) → FOLD_IN_GGA(V, Ys, Z)
FOLD_IN_GGA(X, .(Y, Ys), Z) → U1_GGA(X, Y, Ys, Z, myop_in_gga(X, Y, V))
The TRS R consists of the following rules:
myop_in_gga(a, b, c) → myop_out_gga(a, b, c)
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
myop_in_gga(x1, x2, x3) = myop_in_gga(x1, x2)
a = a
b = b
myop_out_gga(x1, x2, x3) = myop_out_gga(x3)
U1_GGA(x1, x2, x3, x4, x5) = U1_GGA(x3, x5)
FOLD_IN_GGA(x1, x2, x3) = FOLD_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:
FOLD_IN_GGA(X, .(Y, Ys)) → U1_GGA(Ys, myop_in_gga(X, Y))
U1_GGA(Ys, myop_out_gga(V)) → FOLD_IN_GGA(V, Ys)
The TRS R consists of the following rules:
myop_in_gga(a, b) → myop_out_gga(c)
The set Q consists of the following terms:
myop_in_gga(x0, x1)
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:
- FOLD_IN_GGA(X, .(Y, Ys)) → U1_GGA(Ys, myop_in_gga(X, Y))
The graph contains the following edges 2 > 1
- U1_GGA(Ys, myop_out_gga(V)) → FOLD_IN_GGA(V, Ys)
The graph contains the following edges 2 > 1, 1 >= 2