Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(s(0), 0).
p(s(s(X)), s(s(Y))) :- p(s(X), s(Y)).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).

Queries:

plus(g,a,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:

plus_in(s(X), Y, s(Z)) → U2(X, Y, Z, p_in(s(X), U))
p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))
U2(X, Y, Z, p_out(s(X), U)) → U3(X, Y, Z, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U3(X, Y, Z, plus_out(U, Y, Z)) → plus_out(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in(x1, x2, x3)  =  plus_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x4)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
plus_out(x1, x2, x3)  =  plus_out

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:

plus_in(s(X), Y, s(Z)) → U2(X, Y, Z, p_in(s(X), U))
p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))
U2(X, Y, Z, p_out(s(X), U)) → U3(X, Y, Z, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U3(X, Y, Z, plus_out(U, Y, Z)) → plus_out(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in(x1, x2, x3)  =  plus_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x4)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
plus_out(x1, x2, x3)  =  plus_out


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

PLUS_IN(s(X), Y, s(Z)) → U21(X, Y, Z, p_in(s(X), U))
PLUS_IN(s(X), Y, s(Z)) → P_IN(s(X), U)
P_IN(s(s(X)), s(s(Y))) → U11(X, Y, p_in(s(X), s(Y)))
P_IN(s(s(X)), s(s(Y))) → P_IN(s(X), s(Y))
U21(X, Y, Z, p_out(s(X), U)) → U31(X, Y, Z, plus_in(U, Y, Z))
U21(X, Y, Z, p_out(s(X), U)) → PLUS_IN(U, Y, Z)

The TRS R consists of the following rules:

plus_in(s(X), Y, s(Z)) → U2(X, Y, Z, p_in(s(X), U))
p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))
U2(X, Y, Z, p_out(s(X), U)) → U3(X, Y, Z, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U3(X, Y, Z, plus_out(U, Y, Z)) → plus_out(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in(x1, x2, x3)  =  plus_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x4)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
plus_out(x1, x2, x3)  =  plus_out
P_IN(x1, x2)  =  P_IN(x1)
U31(x1, x2, x3, x4)  =  U31(x4)
U21(x1, x2, x3, x4)  =  U21(x4)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)
U11(x1, x2, x3)  =  U11(x3)

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:

PLUS_IN(s(X), Y, s(Z)) → U21(X, Y, Z, p_in(s(X), U))
PLUS_IN(s(X), Y, s(Z)) → P_IN(s(X), U)
P_IN(s(s(X)), s(s(Y))) → U11(X, Y, p_in(s(X), s(Y)))
P_IN(s(s(X)), s(s(Y))) → P_IN(s(X), s(Y))
U21(X, Y, Z, p_out(s(X), U)) → U31(X, Y, Z, plus_in(U, Y, Z))
U21(X, Y, Z, p_out(s(X), U)) → PLUS_IN(U, Y, Z)

The TRS R consists of the following rules:

plus_in(s(X), Y, s(Z)) → U2(X, Y, Z, p_in(s(X), U))
p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))
U2(X, Y, Z, p_out(s(X), U)) → U3(X, Y, Z, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U3(X, Y, Z, plus_out(U, Y, Z)) → plus_out(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in(x1, x2, x3)  =  plus_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x4)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
plus_out(x1, x2, x3)  =  plus_out
P_IN(x1, x2)  =  P_IN(x1)
U31(x1, x2, x3, x4)  =  U31(x4)
U21(x1, x2, x3, x4)  =  U21(x4)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

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

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

P_IN(s(s(X)), s(s(Y))) → P_IN(s(X), s(Y))

The TRS R consists of the following rules:

plus_in(s(X), Y, s(Z)) → U2(X, Y, Z, p_in(s(X), U))
p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))
U2(X, Y, Z, p_out(s(X), U)) → U3(X, Y, Z, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U3(X, Y, Z, plus_out(U, Y, Z)) → plus_out(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in(x1, x2, x3)  =  plus_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x4)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
plus_out(x1, x2, x3)  =  plus_out
P_IN(x1, x2)  =  P_IN(x1)

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

P_IN(s(s(X)), s(s(Y))) → P_IN(s(X), s(Y))

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

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

P_IN(s(s(X))) → P_IN(s(X))

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: 



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

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

PLUS_IN(s(X), Y, s(Z)) → U21(X, Y, Z, p_in(s(X), U))
U21(X, Y, Z, p_out(s(X), U)) → PLUS_IN(U, Y, Z)

The TRS R consists of the following rules:

plus_in(s(X), Y, s(Z)) → U2(X, Y, Z, p_in(s(X), U))
p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))
U2(X, Y, Z, p_out(s(X), U)) → U3(X, Y, Z, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U3(X, Y, Z, plus_out(U, Y, Z)) → plus_out(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in(x1, x2, x3)  =  plus_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4)  =  U2(x4)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
plus_out(x1, x2, x3)  =  plus_out
U21(x1, x2, x3, x4)  =  U21(x4)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

PLUS_IN(s(X), Y, s(Z)) → U21(X, Y, Z, p_in(s(X), U))
U21(X, Y, Z, p_out(s(X), U)) → PLUS_IN(U, Y, Z)

The TRS R consists of the following rules:

p_in(s(s(X)), s(s(Y))) → U1(X, Y, p_in(s(X), s(Y)))
p_in(s(0), 0) → p_out(s(0), 0)
U1(X, Y, p_out(s(X), s(Y))) → p_out(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
p_in(x1, x2)  =  p_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
p_out(x1, x2)  =  p_out(x2)
U21(x1, x2, x3, x4)  =  U21(x4)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

U21(p_out(U)) → PLUS_IN(U)
PLUS_IN(s(X)) → U21(p_in(s(X)))

The TRS R consists of the following rules:

p_in(s(s(X))) → U1(p_in(s(X)))
p_in(s(0)) → p_out(0)
U1(p_out(s(Y))) → p_out(s(s(Y)))

The set Q consists of the following terms:

p_in(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PLUS_IN(s(X)) → U21(p_in(s(X)))

Strictly oriented rules of the TRS R:

p_in(s(0)) → p_out(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1   
POL(PLUS_IN(x1)) = 2 + 2·x1   
POL(U1(x1)) = 2 + x1   
POL(U21(x1)) = x1   
POL(p_in(x1)) = 1 + 2·x1   
POL(p_out(x1)) = 2 + 2·x1   
POL(s(x1)) = 1 + x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ DependencyGraphProof

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

U21(p_out(U)) → PLUS_IN(U)

The TRS R consists of the following rules:

p_in(s(s(X))) → U1(p_in(s(X)))
U1(p_out(s(Y))) → p_out(s(s(Y)))

The set Q consists of the following terms:

p_in(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.