Left Termination of the query pattern
less_in_2(a, g)
w.r.t. the given Prolog program could successfully be proven:
↳ Prolog
↳ PrologToPiTRSProof
Clauses:
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).
Queries:
less(a,g).
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
less_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U1_ag(X, Y, less_in_ag(X, Y))
U1_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2) = less_in_ag(x2)
s(x1) = s(x1)
less_out_ag(x1, x2) = less_out_ag(x1)
U1_ag(x1, x2, x3) = U1_ag(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:
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U1_ag(X, Y, less_in_ag(X, Y))
U1_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2) = less_in_ag(x2)
s(x1) = s(x1)
less_out_ag(x1, x2) = less_out_ag(x1)
U1_ag(x1, x2, x3) = U1_ag(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:
LESS_IN_AG(s(X), s(Y)) → U1_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
The TRS R consists of the following rules:
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U1_ag(X, Y, less_in_ag(X, Y))
U1_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2) = less_in_ag(x2)
s(x1) = s(x1)
less_out_ag(x1, x2) = less_out_ag(x1)
U1_ag(x1, x2, x3) = U1_ag(x3)
U1_AG(x1, x2, x3) = U1_AG(x3)
LESS_IN_AG(x1, x2) = LESS_IN_AG(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:
LESS_IN_AG(s(X), s(Y)) → U1_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
The TRS R consists of the following rules:
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U1_ag(X, Y, less_in_ag(X, Y))
U1_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2) = less_in_ag(x2)
s(x1) = s(x1)
less_out_ag(x1, x2) = less_out_ag(x1)
U1_ag(x1, x2, x3) = U1_ag(x3)
U1_AG(x1, x2, x3) = U1_AG(x3)
LESS_IN_AG(x1, x2) = LESS_IN_AG(x2)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
The TRS R consists of the following rules:
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U1_ag(X, Y, less_in_ag(X, Y))
U1_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
The argument filtering Pi contains the following mapping:
less_in_ag(x1, x2) = less_in_ag(x2)
s(x1) = s(x1)
less_out_ag(x1, x2) = less_out_ag(x1)
U1_ag(x1, x2, x3) = U1_ag(x3)
LESS_IN_AG(x1, x2) = LESS_IN_AG(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:
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
R is empty.
The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
LESS_IN_AG(x1, x2) = LESS_IN_AG(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:
LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
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:
- LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)
The graph contains the following edges 1 > 1