(0) Obligation:

Clauses:

less(0, s(X1)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

less(a,g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

less1(0, s(T4)).
less1(s(0), s(s(T14))).
less1(s(s(T21)), s(s(T20))) :- less1(T21, T20).
less1(s(0), s(s(T33))).
less1(s(s(T40)), s(s(T39))) :- less1(T40, T39).

Queries:

less1(a,g).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
less1_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

less1_in_ag(0, s(T4)) → less1_out_ag(0, s(T4))
less1_in_ag(s(0), s(s(T14))) → less1_out_ag(s(0), s(s(T14)))
less1_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, less1_in_ag(T21, T20))
less1_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, less1_in_ag(T40, T39))
U2_ag(T40, T39, less1_out_ag(T40, T39)) → less1_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, less1_out_ag(T21, T20)) → less1_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
less1_out_ag(x1, x2)  =  less1_out_ag(x1)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
U2_ag(x1, x2, x3)  =  U2_ag(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

less1_in_ag(0, s(T4)) → less1_out_ag(0, s(T4))
less1_in_ag(s(0), s(s(T14))) → less1_out_ag(s(0), s(s(T14)))
less1_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, less1_in_ag(T21, T20))
less1_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, less1_in_ag(T40, T39))
U2_ag(T40, T39, less1_out_ag(T40, T39)) → less1_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, less1_out_ag(T21, T20)) → less1_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
less1_out_ag(x1, x2)  =  less1_out_ag(x1)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
U2_ag(x1, x2, x3)  =  U2_ag(x3)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

LESS1_IN_AG(s(s(T21)), s(s(T20))) → U1_AG(T21, T20, less1_in_ag(T21, T20))
LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)
LESS1_IN_AG(s(s(T40)), s(s(T39))) → U2_AG(T40, T39, less1_in_ag(T40, T39))

The TRS R consists of the following rules:

less1_in_ag(0, s(T4)) → less1_out_ag(0, s(T4))
less1_in_ag(s(0), s(s(T14))) → less1_out_ag(s(0), s(s(T14)))
less1_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, less1_in_ag(T21, T20))
less1_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, less1_in_ag(T40, T39))
U2_ag(T40, T39, less1_out_ag(T40, T39)) → less1_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, less1_out_ag(T21, T20)) → less1_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
less1_out_ag(x1, x2)  =  less1_out_ag(x1)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
U2_AG(x1, x2, x3)  =  U2_AG(x3)

We have to consider all (P,R,Pi)-chains

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

LESS1_IN_AG(s(s(T21)), s(s(T20))) → U1_AG(T21, T20, less1_in_ag(T21, T20))
LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)
LESS1_IN_AG(s(s(T40)), s(s(T39))) → U2_AG(T40, T39, less1_in_ag(T40, T39))

The TRS R consists of the following rules:

less1_in_ag(0, s(T4)) → less1_out_ag(0, s(T4))
less1_in_ag(s(0), s(s(T14))) → less1_out_ag(s(0), s(s(T14)))
less1_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, less1_in_ag(T21, T20))
less1_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, less1_in_ag(T40, T39))
U2_ag(T40, T39, less1_out_ag(T40, T39)) → less1_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, less1_out_ag(T21, T20)) → less1_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
less1_out_ag(x1, x2)  =  less1_out_ag(x1)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
U2_AG(x1, x2, x3)  =  U2_AG(x3)

We have to consider all (P,R,Pi)-chains

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(8) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)

The TRS R consists of the following rules:

less1_in_ag(0, s(T4)) → less1_out_ag(0, s(T4))
less1_in_ag(s(0), s(s(T14))) → less1_out_ag(s(0), s(s(T14)))
less1_in_ag(s(s(T21)), s(s(T20))) → U1_ag(T21, T20, less1_in_ag(T21, T20))
less1_in_ag(s(s(T40)), s(s(T39))) → U2_ag(T40, T39, less1_in_ag(T40, T39))
U2_ag(T40, T39, less1_out_ag(T40, T39)) → less1_out_ag(s(s(T40)), s(s(T39)))
U1_ag(T21, T20, less1_out_ag(T21, T20)) → less1_out_ag(s(s(T21)), s(s(T20)))

The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
less1_out_ag(x1, x2)  =  less1_out_ag(x1)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)

We have to consider all (P,R,Pi)-chains

(9) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(10) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)

We have to consider all (P,R,Pi)-chains

(11) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LESS1_IN_AG(s(s(T20))) → LESS1_IN_AG(T20)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • LESS1_IN_AG(s(s(T20))) → LESS1_IN_AG(T20)
    The graph contains the following edges 1 > 1

(14) YES