(0) Obligation:

Clauses:

div(X1, 0, X2, X3) :- failure(a).
div(0, Y, 0, 0) :- no(zero(Y)).
div(X, Y, s(Z), R) :- ','(no(zero(X)), ','(no(zero(Y)), ','(minus(X, Y, U), ','(!, div(U, Y, Z, R))))).
div(X, Y, X4, X) :- ','(no(zero(X)), no(zero(Y))).
minus(X, 0, X).
minus(s(X), s(Y), Z) :- minus(X, Y, Z).
failure(b).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X5).

Query: div(g,g,a,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

minusA(s(X1), s(X2), X3) :- minusA(X1, X2, X3).
divB(s(X1), s(X2), s(X3), X4) :- minusA(X1, X2, X5).
divB(X1, X2, s(X3), X4) :- ','(minuscC(X1, X2, X5), divB(X5, X2, X3, X4)).

Clauses:

minuscA(X1, 0, X1).
minuscA(s(X1), s(X2), X3) :- minuscA(X1, X2, X3).
divcB(0, X1, 0, 0).
divcB(X1, X2, s(X3), X4) :- ','(minuscC(X1, X2, X5), divcB(X5, X2, X3, X4)).
divcB(X1, X2, X3, X1).
divcB(X1, X2, X3, X1).
minuscC(s(X1), s(X2), X3) :- minuscA(X1, X2, X3).

Afs:

divB(x1, x2, x3, x4)  =  divB(x1, x2)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
divB_in: (b,b,f,f)
minusA_in: (b,b,f)
minuscC_in: (b,b,f)
minuscA_in: (b,b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U2_GGAA(X1, X2, X3, X4, minusA_in_gga(X1, X2, X5))
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → MINUSA_IN_GGA(X1, X2, X5)
MINUSA_IN_GGA(s(X1), s(X2), X3) → U1_GGA(X1, X2, X3, minusA_in_gga(X1, X2, X3))
MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)
DIVB_IN_GGAA(X1, X2, s(X3), X4) → U3_GGAA(X1, X2, X3, X4, minuscC_in_gga(X1, X2, X5))
U3_GGAA(X1, X2, X3, X4, minuscC_out_gga(X1, X2, X5)) → U4_GGAA(X1, X2, X3, X4, divB_in_ggaa(X5, X2, X3, X4))
U3_GGAA(X1, X2, X3, X4, minuscC_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, X2, X3, X4)

The TRS R consists of the following rules:

minuscC_in_gga(s(X1), s(X2), X3) → U9_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
U9_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)

The argument filtering Pi contains the following mapping:
divB_in_ggaa(x1, x2, x3, x4)  =  divB_in_ggaa(x1, x2)
s(x1)  =  s(x1)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minuscC_in_gga(x1, x2, x3)  =  minuscC_in_gga(x1, x2)
U9_gga(x1, x2, x3, x4)  =  U9_gga(x1, x2, x4)
minuscA_in_gga(x1, x2, x3)  =  minuscA_in_gga(x1, x2)
0  =  0
minuscA_out_gga(x1, x2, x3)  =  minuscA_out_gga(x1, x2, x3)
U6_gga(x1, x2, x3, x4)  =  U6_gga(x1, x2, x4)
minuscC_out_gga(x1, x2, x3)  =  minuscC_out_gga(x1, x2, x3)
DIVB_IN_GGAA(x1, x2, x3, x4)  =  DIVB_IN_GGAA(x1, x2)
U2_GGAA(x1, x2, x3, x4, x5)  =  U2_GGAA(x1, x2, x5)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x1, x2, x4)
U3_GGAA(x1, x2, x3, x4, x5)  =  U3_GGAA(x1, x2, x5)
U4_GGAA(x1, x2, x3, x4, x5)  =  U4_GGAA(x1, x2, x5)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → U2_GGAA(X1, X2, X3, X4, minusA_in_gga(X1, X2, X5))
DIVB_IN_GGAA(s(X1), s(X2), s(X3), X4) → MINUSA_IN_GGA(X1, X2, X5)
MINUSA_IN_GGA(s(X1), s(X2), X3) → U1_GGA(X1, X2, X3, minusA_in_gga(X1, X2, X3))
MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)
DIVB_IN_GGAA(X1, X2, s(X3), X4) → U3_GGAA(X1, X2, X3, X4, minuscC_in_gga(X1, X2, X5))
U3_GGAA(X1, X2, X3, X4, minuscC_out_gga(X1, X2, X5)) → U4_GGAA(X1, X2, X3, X4, divB_in_ggaa(X5, X2, X3, X4))
U3_GGAA(X1, X2, X3, X4, minuscC_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, X2, X3, X4)

The TRS R consists of the following rules:

minuscC_in_gga(s(X1), s(X2), X3) → U9_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
U9_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)

The argument filtering Pi contains the following mapping:
divB_in_ggaa(x1, x2, x3, x4)  =  divB_in_ggaa(x1, x2)
s(x1)  =  s(x1)
minusA_in_gga(x1, x2, x3)  =  minusA_in_gga(x1, x2)
minuscC_in_gga(x1, x2, x3)  =  minuscC_in_gga(x1, x2)
U9_gga(x1, x2, x3, x4)  =  U9_gga(x1, x2, x4)
minuscA_in_gga(x1, x2, x3)  =  minuscA_in_gga(x1, x2)
0  =  0
minuscA_out_gga(x1, x2, x3)  =  minuscA_out_gga(x1, x2, x3)
U6_gga(x1, x2, x3, x4)  =  U6_gga(x1, x2, x4)
minuscC_out_gga(x1, x2, x3)  =  minuscC_out_gga(x1, x2, x3)
DIVB_IN_GGAA(x1, x2, x3, x4)  =  DIVB_IN_GGAA(x1, x2)
U2_GGAA(x1, x2, x3, x4, x5)  =  U2_GGAA(x1, x2, x5)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x1, x2, x4)
U3_GGAA(x1, x2, x3, x4, x5)  =  U3_GGAA(x1, x2, x5)
U4_GGAA(x1, x2, x3, x4, x5)  =  U4_GGAA(x1, x2, x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)

The TRS R consists of the following rules:

minuscC_in_gga(s(X1), s(X2), X3) → U9_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
U9_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
minuscC_in_gga(x1, x2, x3)  =  minuscC_in_gga(x1, x2)
U9_gga(x1, x2, x3, x4)  =  U9_gga(x1, x2, x4)
minuscA_in_gga(x1, x2, x3)  =  minuscA_in_gga(x1, x2)
0  =  0
minuscA_out_gga(x1, x2, x3)  =  minuscA_out_gga(x1, x2, x3)
U6_gga(x1, x2, x3, x4)  =  U6_gga(x1, x2, x4)
minuscC_out_gga(x1, x2, x3)  =  minuscC_out_gga(x1, x2, x3)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

MINUSA_IN_GGA(s(X1), s(X2), X3) → MINUSA_IN_GGA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
MINUSA_IN_GGA(x1, x2, x3)  =  MINUSA_IN_GGA(x1, x2)

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

MINUSA_IN_GGA(s(X1), s(X2)) → MINUSA_IN_GGA(X1, X2)

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

(12) 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:

  • MINUSA_IN_GGA(s(X1), s(X2)) → MINUSA_IN_GGA(X1, X2)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

DIVB_IN_GGAA(X1, X2, s(X3), X4) → U3_GGAA(X1, X2, X3, X4, minuscC_in_gga(X1, X2, X5))
U3_GGAA(X1, X2, X3, X4, minuscC_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, X2, X3, X4)

The TRS R consists of the following rules:

minuscC_in_gga(s(X1), s(X2), X3) → U9_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
minuscA_in_gga(X1, 0, X1) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2), X3) → U6_gga(X1, X2, X3, minuscA_in_gga(X1, X2, X3))
U6_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
U9_gga(X1, X2, X3, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
minuscC_in_gga(x1, x2, x3)  =  minuscC_in_gga(x1, x2)
U9_gga(x1, x2, x3, x4)  =  U9_gga(x1, x2, x4)
minuscA_in_gga(x1, x2, x3)  =  minuscA_in_gga(x1, x2)
0  =  0
minuscA_out_gga(x1, x2, x3)  =  minuscA_out_gga(x1, x2, x3)
U6_gga(x1, x2, x3, x4)  =  U6_gga(x1, x2, x4)
minuscC_out_gga(x1, x2, x3)  =  minuscC_out_gga(x1, x2, x3)
DIVB_IN_GGAA(x1, x2, x3, x4)  =  DIVB_IN_GGAA(x1, x2)
U3_GGAA(x1, x2, x3, x4, x5)  =  U3_GGAA(x1, x2, x5)

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

(15) PiDPToQDPProof (SOUND transformation)

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

(16) Obligation:

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

DIVB_IN_GGAA(X1, X2) → U3_GGAA(X1, X2, minuscC_in_gga(X1, X2))
U3_GGAA(X1, X2, minuscC_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, X2)

The TRS R consists of the following rules:

minuscC_in_gga(s(X1), s(X2)) → U9_gga(X1, X2, minuscA_in_gga(X1, X2))
minuscA_in_gga(X1, 0) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2)) → U6_gga(X1, X2, minuscA_in_gga(X1, X2))
U6_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
U9_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)

The set Q consists of the following terms:

minuscC_in_gga(x0, x1)
minuscA_in_gga(x0, x1)
U6_gga(x0, x1, x2)
U9_gga(x0, x1, x2)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


U3_GGAA(X1, X2, minuscC_out_gga(X1, X2, X5)) → DIVB_IN_GGAA(X5, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIVB_IN_GGAA(x1, x2)) = 1 + x1 + x2   
POL(U3_GGAA(x1, x2, x3)) = 1 + x3   
POL(U6_gga(x1, x2, x3)) = x3   
POL(U9_gga(x1, x2, x3)) = 1 + x2 + x3   
POL(minuscA_in_gga(x1, x2)) = 1 + x1   
POL(minuscA_out_gga(x1, x2, x3)) = 1 + x3   
POL(minuscC_in_gga(x1, x2)) = x1 + x2   
POL(minuscC_out_gga(x1, x2, x3)) = 1 + x2 + x3   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

minuscC_in_gga(s(X1), s(X2)) → U9_gga(X1, X2, minuscA_in_gga(X1, X2))
minuscA_in_gga(X1, 0) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2)) → U6_gga(X1, X2, minuscA_in_gga(X1, X2))
U9_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)
U6_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)

(18) Obligation:

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

DIVB_IN_GGAA(X1, X2) → U3_GGAA(X1, X2, minuscC_in_gga(X1, X2))

The TRS R consists of the following rules:

minuscC_in_gga(s(X1), s(X2)) → U9_gga(X1, X2, minuscA_in_gga(X1, X2))
minuscA_in_gga(X1, 0) → minuscA_out_gga(X1, 0, X1)
minuscA_in_gga(s(X1), s(X2)) → U6_gga(X1, X2, minuscA_in_gga(X1, X2))
U6_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscA_out_gga(s(X1), s(X2), X3)
U9_gga(X1, X2, minuscA_out_gga(X1, X2, X3)) → minuscC_out_gga(s(X1), s(X2), X3)

The set Q consists of the following terms:

minuscC_in_gga(x0, x1)
minuscA_in_gga(x0, x1)
U6_gga(x0, x1, x2)
U9_gga(x0, x1, x2)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(20) TRUE