(0) Obligation:

Clauses:

less(0, Y) :- ','(!, =(Y, s(X2))).
less(X, Y) :- ','(p(X, X1), ','(p(Y, Y1), less(X1, Y1))).
p(0, 0).
p(s(X), X).
=(X, X).

Query: less(g,a)

(1) BuiltinConflictTransformerProof (EQUIVALENT transformation)

Renamed defined predicates conflicting with built-in predicates [PROLOG].

(2) Obligation:

Clauses:

less(0, Y) :- ','(!, user_defined_=(Y, s(X2))).
less(X, Y) :- ','(p(X, X1), ','(p(Y, Y1), less(X1, Y1))).
p(0, 0).
p(s(X), X).
user_defined_=(X, X).

Query: less(g,a)

(3) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

(4) Obligation:

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

lessA_in_ga(0, s(T11)) → lessA_out_ga(0, s(T11))
lessA_in_ga(s(T19), 0) → U1_ga(T19, lessB_in_g(T19))
lessB_in_g(s(T25)) → U3_g(T25, lessB_in_g(T25))
U3_g(T25, lessB_out_g(T25)) → lessB_out_g(s(T25))
U1_ga(T19, lessB_out_g(T19)) → lessA_out_ga(s(T19), 0)
lessA_in_ga(s(T19), s(T29)) → U2_ga(T19, T29, lessA_in_ga(T19, T29))
U2_ga(T19, T29, lessA_out_ga(T19, T29)) → lessA_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessA_in_ga(x1, x2)  =  lessA_in_ga(x1)
0  =  0
lessA_out_ga(x1, x2)  =  lessA_out_ga(x1)
s(x1)  =  s(x1)
U1_ga(x1, x2)  =  U1_ga(x1, x2)
lessB_in_g(x1)  =  lessB_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
lessB_out_g(x1)  =  lessB_out_g(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, 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:

LESSA_IN_GA(s(T19), 0) → U1_GA(T19, lessB_in_g(T19))
LESSA_IN_GA(s(T19), 0) → LESSB_IN_G(T19)
LESSB_IN_G(s(T25)) → U3_G(T25, lessB_in_g(T25))
LESSB_IN_G(s(T25)) → LESSB_IN_G(T25)
LESSA_IN_GA(s(T19), s(T29)) → U2_GA(T19, T29, lessA_in_ga(T19, T29))
LESSA_IN_GA(s(T19), s(T29)) → LESSA_IN_GA(T19, T29)

The TRS R consists of the following rules:

lessA_in_ga(0, s(T11)) → lessA_out_ga(0, s(T11))
lessA_in_ga(s(T19), 0) → U1_ga(T19, lessB_in_g(T19))
lessB_in_g(s(T25)) → U3_g(T25, lessB_in_g(T25))
U3_g(T25, lessB_out_g(T25)) → lessB_out_g(s(T25))
U1_ga(T19, lessB_out_g(T19)) → lessA_out_ga(s(T19), 0)
lessA_in_ga(s(T19), s(T29)) → U2_ga(T19, T29, lessA_in_ga(T19, T29))
U2_ga(T19, T29, lessA_out_ga(T19, T29)) → lessA_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessA_in_ga(x1, x2)  =  lessA_in_ga(x1)
0  =  0
lessA_out_ga(x1, x2)  =  lessA_out_ga(x1)
s(x1)  =  s(x1)
U1_ga(x1, x2)  =  U1_ga(x1, x2)
lessB_in_g(x1)  =  lessB_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
lessB_out_g(x1)  =  lessB_out_g(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
LESSA_IN_GA(x1, x2)  =  LESSA_IN_GA(x1)
U1_GA(x1, x2)  =  U1_GA(x1, x2)
LESSB_IN_G(x1)  =  LESSB_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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

(6) Obligation:

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

LESSA_IN_GA(s(T19), 0) → U1_GA(T19, lessB_in_g(T19))
LESSA_IN_GA(s(T19), 0) → LESSB_IN_G(T19)
LESSB_IN_G(s(T25)) → U3_G(T25, lessB_in_g(T25))
LESSB_IN_G(s(T25)) → LESSB_IN_G(T25)
LESSA_IN_GA(s(T19), s(T29)) → U2_GA(T19, T29, lessA_in_ga(T19, T29))
LESSA_IN_GA(s(T19), s(T29)) → LESSA_IN_GA(T19, T29)

The TRS R consists of the following rules:

lessA_in_ga(0, s(T11)) → lessA_out_ga(0, s(T11))
lessA_in_ga(s(T19), 0) → U1_ga(T19, lessB_in_g(T19))
lessB_in_g(s(T25)) → U3_g(T25, lessB_in_g(T25))
U3_g(T25, lessB_out_g(T25)) → lessB_out_g(s(T25))
U1_ga(T19, lessB_out_g(T19)) → lessA_out_ga(s(T19), 0)
lessA_in_ga(s(T19), s(T29)) → U2_ga(T19, T29, lessA_in_ga(T19, T29))
U2_ga(T19, T29, lessA_out_ga(T19, T29)) → lessA_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessA_in_ga(x1, x2)  =  lessA_in_ga(x1)
0  =  0
lessA_out_ga(x1, x2)  =  lessA_out_ga(x1)
s(x1)  =  s(x1)
U1_ga(x1, x2)  =  U1_ga(x1, x2)
lessB_in_g(x1)  =  lessB_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
lessB_out_g(x1)  =  lessB_out_g(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
LESSA_IN_GA(x1, x2)  =  LESSA_IN_GA(x1)
U1_GA(x1, x2)  =  U1_GA(x1, x2)
LESSB_IN_G(x1)  =  LESSB_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

LESSB_IN_G(s(T25)) → LESSB_IN_G(T25)

The TRS R consists of the following rules:

lessA_in_ga(0, s(T11)) → lessA_out_ga(0, s(T11))
lessA_in_ga(s(T19), 0) → U1_ga(T19, lessB_in_g(T19))
lessB_in_g(s(T25)) → U3_g(T25, lessB_in_g(T25))
U3_g(T25, lessB_out_g(T25)) → lessB_out_g(s(T25))
U1_ga(T19, lessB_out_g(T19)) → lessA_out_ga(s(T19), 0)
lessA_in_ga(s(T19), s(T29)) → U2_ga(T19, T29, lessA_in_ga(T19, T29))
U2_ga(T19, T29, lessA_out_ga(T19, T29)) → lessA_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessA_in_ga(x1, x2)  =  lessA_in_ga(x1)
0  =  0
lessA_out_ga(x1, x2)  =  lessA_out_ga(x1)
s(x1)  =  s(x1)
U1_ga(x1, x2)  =  U1_ga(x1, x2)
lessB_in_g(x1)  =  lessB_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
lessB_out_g(x1)  =  lessB_out_g(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
LESSB_IN_G(x1)  =  LESSB_IN_G(x1)

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

LESSB_IN_G(s(T25)) → LESSB_IN_G(T25)

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

(12) PiDPToQDPProof (EQUIVALENT transformation)

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

(13) Obligation:

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

LESSB_IN_G(s(T25)) → LESSB_IN_G(T25)

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

(14) 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:

  • LESSB_IN_G(s(T25)) → LESSB_IN_G(T25)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

LESSA_IN_GA(s(T19), s(T29)) → LESSA_IN_GA(T19, T29)

The TRS R consists of the following rules:

lessA_in_ga(0, s(T11)) → lessA_out_ga(0, s(T11))
lessA_in_ga(s(T19), 0) → U1_ga(T19, lessB_in_g(T19))
lessB_in_g(s(T25)) → U3_g(T25, lessB_in_g(T25))
U3_g(T25, lessB_out_g(T25)) → lessB_out_g(s(T25))
U1_ga(T19, lessB_out_g(T19)) → lessA_out_ga(s(T19), 0)
lessA_in_ga(s(T19), s(T29)) → U2_ga(T19, T29, lessA_in_ga(T19, T29))
U2_ga(T19, T29, lessA_out_ga(T19, T29)) → lessA_out_ga(s(T19), s(T29))

The argument filtering Pi contains the following mapping:
lessA_in_ga(x1, x2)  =  lessA_in_ga(x1)
0  =  0
lessA_out_ga(x1, x2)  =  lessA_out_ga(x1)
s(x1)  =  s(x1)
U1_ga(x1, x2)  =  U1_ga(x1, x2)
lessB_in_g(x1)  =  lessB_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
lessB_out_g(x1)  =  lessB_out_g(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
LESSA_IN_GA(x1, x2)  =  LESSA_IN_GA(x1)

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

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

LESSA_IN_GA(s(T19), s(T29)) → LESSA_IN_GA(T19, T29)

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

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

LESSA_IN_GA(s(T19)) → LESSA_IN_GA(T19)

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

(21) 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:

  • LESSA_IN_GA(s(T19)) → LESSA_IN_GA(T19)
    The graph contains the following edges 1 > 1

(22) YES