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(X), X).
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)) → U1(X, Y, Z, p_in(s(X), U))
p_in(s(X), X) → p_out(s(X), X)
U1(X, Y, Z, p_out(s(X), U)) → U2(X, Y, Z, U, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U2(X, Y, Z, U, 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)
U1(x1, x2, x3, x4)  =  U1(x4)
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
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)) → U1(X, Y, Z, p_in(s(X), U))
p_in(s(X), X) → p_out(s(X), X)
U1(X, Y, Z, p_out(s(X), U)) → U2(X, Y, Z, U, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U2(X, Y, Z, U, 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)
U1(x1, x2, x3, x4)  =  U1(x4)
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
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)) → U11(X, Y, Z, p_in(s(X), U))
PLUS_IN(s(X), Y, s(Z)) → P_IN(s(X), U)
U11(X, Y, Z, p_out(s(X), U)) → U21(X, Y, Z, U, plus_in(U, Y, Z))
U11(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)) → U1(X, Y, Z, p_in(s(X), U))
p_in(s(X), X) → p_out(s(X), X)
U1(X, Y, Z, p_out(s(X), U)) → U2(X, Y, Z, U, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U2(X, Y, Z, U, 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)
U1(x1, x2, x3, x4)  =  U1(x4)
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
plus_out(x1, x2, x3)  =  plus_out
P_IN(x1, x2)  =  P_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)
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:

PLUS_IN(s(X), Y, s(Z)) → U11(X, Y, Z, p_in(s(X), U))
PLUS_IN(s(X), Y, s(Z)) → P_IN(s(X), U)
U11(X, Y, Z, p_out(s(X), U)) → U21(X, Y, Z, U, plus_in(U, Y, Z))
U11(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)) → U1(X, Y, Z, p_in(s(X), U))
p_in(s(X), X) → p_out(s(X), X)
U1(X, Y, Z, p_out(s(X), U)) → U2(X, Y, Z, U, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U2(X, Y, Z, U, 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)
U1(x1, x2, x3, x4)  =  U1(x4)
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
plus_out(x1, x2, x3)  =  plus_out
P_IN(x1, x2)  =  P_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)
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 2 less nodes.

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

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

PLUS_IN(s(X), Y, s(Z)) → U11(X, Y, Z, p_in(s(X), U))
U11(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)) → U1(X, Y, Z, p_in(s(X), U))
p_in(s(X), X) → p_out(s(X), X)
U1(X, Y, Z, p_out(s(X), U)) → U2(X, Y, Z, U, plus_in(U, Y, Z))
plus_in(0, Y, Y) → plus_out(0, Y, Y)
U2(X, Y, Z, U, 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)
U1(x1, x2, x3, x4)  =  U1(x4)
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
plus_out(x1, x2, x3)  =  plus_out
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x4)

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:

PLUS_IN(s(X), Y, s(Z)) → U11(X, Y, Z, p_in(s(X), U))
U11(X, Y, Z, p_out(s(X), U)) → PLUS_IN(U, Y, Z)

The TRS R consists of the following rules:

p_in(s(X), X) → p_out(s(X), X)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
PLUS_IN(x1, x2, x3)  =  PLUS_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x4)

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
                      ↳ RuleRemovalProof

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

U11(p_out(U)) → PLUS_IN(U)
PLUS_IN(s(X)) → U11(p_in(s(X)))

The TRS R consists of the following rules:

p_in(s(X)) → p_out(X)

The set Q consists of the following terms:

p_in(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)) → U11(p_in(s(X)))

Strictly oriented rules of the TRS R:

p_in(s(X)) → p_out(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(PLUS_IN(x1)) = 2 + 2·x1   
POL(U11(x1)) = x1   
POL(p_in(x1)) = 1 + 2·x1   
POL(p_out(x1)) = 2 + 2·x1   
POL(s(x1)) = 1 + 2·x1   



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

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

U11(p_out(U)) → PLUS_IN(U)

R is empty.
The set Q consists of the following terms:

p_in(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.