Left Termination of the query pattern div_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

div(X, Y, Z) :- quot(X, Y, Y, Z).
quot(0, s(Y), s(Z), 0).
quot(s(X), s(Y), Z, U) :- quot(X, Y, Z, U).
quot(X, 0, s(Z), s(U)) :- quot(X, s(Z), s(Z), U).

Queries:

div(g,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:

div_in(X, Y, Z) → U1(X, Y, Z, quot_in(X, Y, Y, Z))
quot_in(X, 0, s(Z), s(U)) → U3(X, Z, U, quot_in(X, s(Z), s(Z), U))
quot_in(s(X), s(Y), Z, U) → U2(X, Y, Z, U, quot_in(X, Y, Z, U))
quot_in(0, s(Y), s(Z), 0) → quot_out(0, s(Y), s(Z), 0)
U2(X, Y, Z, U, quot_out(X, Y, Z, U)) → quot_out(s(X), s(Y), Z, U)
U3(X, Z, U, quot_out(X, s(Z), s(Z), U)) → quot_out(X, 0, s(Z), s(U))
U1(X, Y, Z, quot_out(X, Y, Y, Z)) → div_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
div_in(x1, x2, x3)  =  div_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
quot_in(x1, x2, x3, x4)  =  quot_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
quot_out(x1, x2, x3, x4)  =  quot_out(x4)
div_out(x1, x2, x3)  =  div_out(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:

div_in(X, Y, Z) → U1(X, Y, Z, quot_in(X, Y, Y, Z))
quot_in(X, 0, s(Z), s(U)) → U3(X, Z, U, quot_in(X, s(Z), s(Z), U))
quot_in(s(X), s(Y), Z, U) → U2(X, Y, Z, U, quot_in(X, Y, Z, U))
quot_in(0, s(Y), s(Z), 0) → quot_out(0, s(Y), s(Z), 0)
U2(X, Y, Z, U, quot_out(X, Y, Z, U)) → quot_out(s(X), s(Y), Z, U)
U3(X, Z, U, quot_out(X, s(Z), s(Z), U)) → quot_out(X, 0, s(Z), s(U))
U1(X, Y, Z, quot_out(X, Y, Y, Z)) → div_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
div_in(x1, x2, x3)  =  div_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
quot_in(x1, x2, x3, x4)  =  quot_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
quot_out(x1, x2, x3, x4)  =  quot_out(x4)
div_out(x1, x2, x3)  =  div_out(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:

DIV_IN(X, Y, Z) → U11(X, Y, Z, quot_in(X, Y, Y, Z))
DIV_IN(X, Y, Z) → QUOT_IN(X, Y, Y, Z)
QUOT_IN(X, 0, s(Z), s(U)) → U31(X, Z, U, quot_in(X, s(Z), s(Z), U))
QUOT_IN(X, 0, s(Z), s(U)) → QUOT_IN(X, s(Z), s(Z), U)
QUOT_IN(s(X), s(Y), Z, U) → U21(X, Y, Z, U, quot_in(X, Y, Z, U))
QUOT_IN(s(X), s(Y), Z, U) → QUOT_IN(X, Y, Z, U)

The TRS R consists of the following rules:

div_in(X, Y, Z) → U1(X, Y, Z, quot_in(X, Y, Y, Z))
quot_in(X, 0, s(Z), s(U)) → U3(X, Z, U, quot_in(X, s(Z), s(Z), U))
quot_in(s(X), s(Y), Z, U) → U2(X, Y, Z, U, quot_in(X, Y, Z, U))
quot_in(0, s(Y), s(Z), 0) → quot_out(0, s(Y), s(Z), 0)
U2(X, Y, Z, U, quot_out(X, Y, Z, U)) → quot_out(s(X), s(Y), Z, U)
U3(X, Z, U, quot_out(X, s(Z), s(Z), U)) → quot_out(X, 0, s(Z), s(U))
U1(X, Y, Z, quot_out(X, Y, Y, Z)) → div_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
div_in(x1, x2, x3)  =  div_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
quot_in(x1, x2, x3, x4)  =  quot_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
quot_out(x1, x2, x3, x4)  =  quot_out(x4)
div_out(x1, x2, x3)  =  div_out(x3)
QUOT_IN(x1, x2, x3, x4)  =  QUOT_IN(x1, x2, x3)
U31(x1, x2, x3, x4)  =  U31(x4)
DIV_IN(x1, x2, x3)  =  DIV_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)

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:

DIV_IN(X, Y, Z) → U11(X, Y, Z, quot_in(X, Y, Y, Z))
DIV_IN(X, Y, Z) → QUOT_IN(X, Y, Y, Z)
QUOT_IN(X, 0, s(Z), s(U)) → U31(X, Z, U, quot_in(X, s(Z), s(Z), U))
QUOT_IN(X, 0, s(Z), s(U)) → QUOT_IN(X, s(Z), s(Z), U)
QUOT_IN(s(X), s(Y), Z, U) → U21(X, Y, Z, U, quot_in(X, Y, Z, U))
QUOT_IN(s(X), s(Y), Z, U) → QUOT_IN(X, Y, Z, U)

The TRS R consists of the following rules:

div_in(X, Y, Z) → U1(X, Y, Z, quot_in(X, Y, Y, Z))
quot_in(X, 0, s(Z), s(U)) → U3(X, Z, U, quot_in(X, s(Z), s(Z), U))
quot_in(s(X), s(Y), Z, U) → U2(X, Y, Z, U, quot_in(X, Y, Z, U))
quot_in(0, s(Y), s(Z), 0) → quot_out(0, s(Y), s(Z), 0)
U2(X, Y, Z, U, quot_out(X, Y, Z, U)) → quot_out(s(X), s(Y), Z, U)
U3(X, Z, U, quot_out(X, s(Z), s(Z), U)) → quot_out(X, 0, s(Z), s(U))
U1(X, Y, Z, quot_out(X, Y, Y, Z)) → div_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
div_in(x1, x2, x3)  =  div_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
quot_in(x1, x2, x3, x4)  =  quot_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
quot_out(x1, x2, x3, x4)  =  quot_out(x4)
div_out(x1, x2, x3)  =  div_out(x3)
QUOT_IN(x1, x2, x3, x4)  =  QUOT_IN(x1, x2, x3)
U31(x1, x2, x3, x4)  =  U31(x4)
DIV_IN(x1, x2, x3)  =  DIV_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 4 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

QUOT_IN(s(X), s(Y), Z, U) → QUOT_IN(X, Y, Z, U)
QUOT_IN(X, 0, s(Z), s(U)) → QUOT_IN(X, s(Z), s(Z), U)

The TRS R consists of the following rules:

div_in(X, Y, Z) → U1(X, Y, Z, quot_in(X, Y, Y, Z))
quot_in(X, 0, s(Z), s(U)) → U3(X, Z, U, quot_in(X, s(Z), s(Z), U))
quot_in(s(X), s(Y), Z, U) → U2(X, Y, Z, U, quot_in(X, Y, Z, U))
quot_in(0, s(Y), s(Z), 0) → quot_out(0, s(Y), s(Z), 0)
U2(X, Y, Z, U, quot_out(X, Y, Z, U)) → quot_out(s(X), s(Y), Z, U)
U3(X, Z, U, quot_out(X, s(Z), s(Z), U)) → quot_out(X, 0, s(Z), s(U))
U1(X, Y, Z, quot_out(X, Y, Y, Z)) → div_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
div_in(x1, x2, x3)  =  div_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
quot_in(x1, x2, x3, x4)  =  quot_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4)  =  U3(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
quot_out(x1, x2, x3, x4)  =  quot_out(x4)
div_out(x1, x2, x3)  =  div_out(x3)
QUOT_IN(x1, x2, x3, x4)  =  QUOT_IN(x1, x2, 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:

QUOT_IN(s(X), s(Y), Z, U) → QUOT_IN(X, Y, Z, U)
QUOT_IN(X, 0, s(Z), s(U)) → QUOT_IN(X, s(Z), s(Z), U)

R is empty.
The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
QUOT_IN(x1, x2, x3, x4)  =  QUOT_IN(x1, x2, 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
                      ↳ QDPSizeChangeProof

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

QUOT_IN(X, 0, s(Z)) → QUOT_IN(X, s(Z), s(Z))
QUOT_IN(s(X), s(Y), Z) → QUOT_IN(X, Y, Z)

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: