Left Termination of the query pattern p_in_4(g, g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(M, N, s(R), RES) :- p(M, R, N, RES).
p(M, s(N), R, RES) :- p(R, N, M, RES).
p(M, X, X1, M).

Queries:

p(g,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:

p_in(M, X, X1, M) → p_out(M, X, X1, M)
p_in(M, s(N), R, RES) → U2(M, N, R, RES, p_in(R, N, M, RES))
p_in(M, N, s(R), RES) → U1(M, N, R, RES, p_in(M, R, N, RES))
U1(M, N, R, RES, p_out(M, R, N, RES)) → p_out(M, N, s(R), RES)
U2(M, N, R, RES, p_out(R, N, M, RES)) → p_out(M, s(N), R, RES)

The argument filtering Pi contains the following mapping:
p_in(x1, x2, x3, x4)  =  p_in(x1, x2, x3)
p_out(x1, x2, x3, x4)  =  p_out(x4)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)

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:

p_in(M, X, X1, M) → p_out(M, X, X1, M)
p_in(M, s(N), R, RES) → U2(M, N, R, RES, p_in(R, N, M, RES))
p_in(M, N, s(R), RES) → U1(M, N, R, RES, p_in(M, R, N, RES))
U1(M, N, R, RES, p_out(M, R, N, RES)) → p_out(M, N, s(R), RES)
U2(M, N, R, RES, p_out(R, N, M, RES)) → p_out(M, s(N), R, RES)

The argument filtering Pi contains the following mapping:
p_in(x1, x2, x3, x4)  =  p_in(x1, x2, x3)
p_out(x1, x2, x3, x4)  =  p_out(x4)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)


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

P_IN(M, s(N), R, RES) → U21(M, N, R, RES, p_in(R, N, M, RES))
P_IN(M, s(N), R, RES) → P_IN(R, N, M, RES)
P_IN(M, N, s(R), RES) → U11(M, N, R, RES, p_in(M, R, N, RES))
P_IN(M, N, s(R), RES) → P_IN(M, R, N, RES)

The TRS R consists of the following rules:

p_in(M, X, X1, M) → p_out(M, X, X1, M)
p_in(M, s(N), R, RES) → U2(M, N, R, RES, p_in(R, N, M, RES))
p_in(M, N, s(R), RES) → U1(M, N, R, RES, p_in(M, R, N, RES))
U1(M, N, R, RES, p_out(M, R, N, RES)) → p_out(M, N, s(R), RES)
U2(M, N, R, RES, p_out(R, N, M, RES)) → p_out(M, s(N), R, RES)

The argument filtering Pi contains the following mapping:
p_in(x1, x2, x3, x4)  =  p_in(x1, x2, x3)
p_out(x1, x2, x3, x4)  =  p_out(x4)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)
P_IN(x1, x2, x3, x4)  =  P_IN(x1, x2, 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:

P_IN(M, s(N), R, RES) → U21(M, N, R, RES, p_in(R, N, M, RES))
P_IN(M, s(N), R, RES) → P_IN(R, N, M, RES)
P_IN(M, N, s(R), RES) → U11(M, N, R, RES, p_in(M, R, N, RES))
P_IN(M, N, s(R), RES) → P_IN(M, R, N, RES)

The TRS R consists of the following rules:

p_in(M, X, X1, M) → p_out(M, X, X1, M)
p_in(M, s(N), R, RES) → U2(M, N, R, RES, p_in(R, N, M, RES))
p_in(M, N, s(R), RES) → U1(M, N, R, RES, p_in(M, R, N, RES))
U1(M, N, R, RES, p_out(M, R, N, RES)) → p_out(M, N, s(R), RES)
U2(M, N, R, RES, p_out(R, N, M, RES)) → p_out(M, s(N), R, RES)

The argument filtering Pi contains the following mapping:
p_in(x1, x2, x3, x4)  =  p_in(x1, x2, x3)
p_out(x1, x2, x3, x4)  =  p_out(x4)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)
P_IN(x1, x2, x3, x4)  =  P_IN(x1, x2, x3)

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:

P_IN(M, N, s(R), RES) → P_IN(M, R, N, RES)
P_IN(M, s(N), R, RES) → P_IN(R, N, M, RES)

The TRS R consists of the following rules:

p_in(M, X, X1, M) → p_out(M, X, X1, M)
p_in(M, s(N), R, RES) → U2(M, N, R, RES, p_in(R, N, M, RES))
p_in(M, N, s(R), RES) → U1(M, N, R, RES, p_in(M, R, N, RES))
U1(M, N, R, RES, p_out(M, R, N, RES)) → p_out(M, N, s(R), RES)
U2(M, N, R, RES, p_out(R, N, M, RES)) → p_out(M, s(N), R, RES)

The argument filtering Pi contains the following mapping:
p_in(x1, x2, x3, x4)  =  p_in(x1, x2, x3)
p_out(x1, x2, x3, x4)  =  p_out(x4)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
P_IN(x1, x2, x3, x4)  =  P_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:

P_IN(M, N, s(R), RES) → P_IN(M, R, N, RES)
P_IN(M, s(N), R, RES) → P_IN(R, N, M, RES)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
P_IN(x1, x2, x3, x4)  =  P_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:

P_IN(M, s(N), R) → P_IN(R, N, M)
P_IN(M, N, s(R)) → P_IN(M, R, N)

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: