Left Termination of the query pattern
tc_in_2(g, a)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
p(a, b).
p(b, c).
tc(X, X).
tc(X, Y) :- ','(p(X, Z), tc(Z, Y)).
Queries:
tc(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:
tc_in(X, Y) → U1(X, Y, p_in(X, Z))
p_in(b, c) → p_out(b, c)
p_in(a, b) → p_out(a, b)
U1(X, Y, p_out(X, Z)) → U2(X, Y, Z, tc_in(Z, Y))
tc_in(X, X) → tc_out(X, X)
U2(X, Y, Z, tc_out(Z, Y)) → tc_out(X, Y)
The argument filtering Pi contains the following mapping:
tc_in(x1, x2) = tc_in(x1)
U1(x1, x2, x3) = U1(x3)
p_in(x1, x2) = p_in(x1)
b = b
c = c
p_out(x1, x2) = p_out(x2)
a = a
U2(x1, x2, x3, x4) = U2(x4)
tc_out(x1, x2) = tc_out(x2)
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:
tc_in(X, Y) → U1(X, Y, p_in(X, Z))
p_in(b, c) → p_out(b, c)
p_in(a, b) → p_out(a, b)
U1(X, Y, p_out(X, Z)) → U2(X, Y, Z, tc_in(Z, Y))
tc_in(X, X) → tc_out(X, X)
U2(X, Y, Z, tc_out(Z, Y)) → tc_out(X, Y)
The argument filtering Pi contains the following mapping:
tc_in(x1, x2) = tc_in(x1)
U1(x1, x2, x3) = U1(x3)
p_in(x1, x2) = p_in(x1)
b = b
c = c
p_out(x1, x2) = p_out(x2)
a = a
U2(x1, x2, x3, x4) = U2(x4)
tc_out(x1, x2) = tc_out(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:
TC_IN(X, Y) → U11(X, Y, p_in(X, Z))
TC_IN(X, Y) → P_IN(X, Z)
U11(X, Y, p_out(X, Z)) → U21(X, Y, Z, tc_in(Z, Y))
U11(X, Y, p_out(X, Z)) → TC_IN(Z, Y)
The TRS R consists of the following rules:
tc_in(X, Y) → U1(X, Y, p_in(X, Z))
p_in(b, c) → p_out(b, c)
p_in(a, b) → p_out(a, b)
U1(X, Y, p_out(X, Z)) → U2(X, Y, Z, tc_in(Z, Y))
tc_in(X, X) → tc_out(X, X)
U2(X, Y, Z, tc_out(Z, Y)) → tc_out(X, Y)
The argument filtering Pi contains the following mapping:
tc_in(x1, x2) = tc_in(x1)
U1(x1, x2, x3) = U1(x3)
p_in(x1, x2) = p_in(x1)
b = b
c = c
p_out(x1, x2) = p_out(x2)
a = a
U2(x1, x2, x3, x4) = U2(x4)
tc_out(x1, x2) = tc_out(x2)
P_IN(x1, x2) = P_IN(x1)
U21(x1, x2, x3, x4) = U21(x4)
TC_IN(x1, x2) = TC_IN(x1)
U11(x1, x2, x3) = U11(x3)
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:
TC_IN(X, Y) → U11(X, Y, p_in(X, Z))
TC_IN(X, Y) → P_IN(X, Z)
U11(X, Y, p_out(X, Z)) → U21(X, Y, Z, tc_in(Z, Y))
U11(X, Y, p_out(X, Z)) → TC_IN(Z, Y)
The TRS R consists of the following rules:
tc_in(X, Y) → U1(X, Y, p_in(X, Z))
p_in(b, c) → p_out(b, c)
p_in(a, b) → p_out(a, b)
U1(X, Y, p_out(X, Z)) → U2(X, Y, Z, tc_in(Z, Y))
tc_in(X, X) → tc_out(X, X)
U2(X, Y, Z, tc_out(Z, Y)) → tc_out(X, Y)
The argument filtering Pi contains the following mapping:
tc_in(x1, x2) = tc_in(x1)
U1(x1, x2, x3) = U1(x3)
p_in(x1, x2) = p_in(x1)
b = b
c = c
p_out(x1, x2) = p_out(x2)
a = a
U2(x1, x2, x3, x4) = U2(x4)
tc_out(x1, x2) = tc_out(x2)
P_IN(x1, x2) = P_IN(x1)
U21(x1, x2, x3, x4) = U21(x4)
TC_IN(x1, x2) = TC_IN(x1)
U11(x1, x2, x3) = U11(x3)
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:
TC_IN(X, Y) → U11(X, Y, p_in(X, Z))
U11(X, Y, p_out(X, Z)) → TC_IN(Z, Y)
The TRS R consists of the following rules:
tc_in(X, Y) → U1(X, Y, p_in(X, Z))
p_in(b, c) → p_out(b, c)
p_in(a, b) → p_out(a, b)
U1(X, Y, p_out(X, Z)) → U2(X, Y, Z, tc_in(Z, Y))
tc_in(X, X) → tc_out(X, X)
U2(X, Y, Z, tc_out(Z, Y)) → tc_out(X, Y)
The argument filtering Pi contains the following mapping:
tc_in(x1, x2) = tc_in(x1)
U1(x1, x2, x3) = U1(x3)
p_in(x1, x2) = p_in(x1)
b = b
c = c
p_out(x1, x2) = p_out(x2)
a = a
U2(x1, x2, x3, x4) = U2(x4)
tc_out(x1, x2) = tc_out(x2)
TC_IN(x1, x2) = TC_IN(x1)
U11(x1, x2, x3) = U11(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
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
TC_IN(X, Y) → U11(X, Y, p_in(X, Z))
U11(X, Y, p_out(X, Z)) → TC_IN(Z, Y)
The TRS R consists of the following rules:
p_in(b, c) → p_out(b, c)
p_in(a, b) → p_out(a, b)
The argument filtering Pi contains the following mapping:
p_in(x1, x2) = p_in(x1)
b = b
c = c
p_out(x1, x2) = p_out(x2)
a = a
TC_IN(x1, x2) = TC_IN(x1)
U11(x1, x2, x3) = U11(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
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
TC_IN(X) → U11(p_in(X))
U11(p_out(Z)) → TC_IN(Z)
The TRS R consists of the following rules:
p_in(b) → p_out(c)
p_in(a) → p_out(b)
The set Q consists of the following terms:
p_in(x0)
We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
The following rules are removed from R:
p_in(b) → p_out(c)
p_in(a) → p_out(b)
Used ordering: POLO with Polynomial interpretation [25]:
POL(TC_IN(x1)) = 2·x1
POL(U11(x1)) = 2·x1
POL(a) = 1
POL(b) = 1
POL(c) = 0
POL(p_in(x1)) = x1
POL(p_out(x1)) = x1
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
TC_IN(X) → U11(p_in(X))
U11(p_out(Z)) → TC_IN(Z)
R is empty.
The set Q consists of the following terms:
p_in(x0)
We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.