Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(.(X, [])).
p(.(s(s(X)), .(Y, Xs))) :- ','(p(.(X, .(Y, Xs))), p(.(s(s(s(s(Y)))), Xs))).
p(.(0, Xs)) :- p(Xs).

Queries:

p(g).

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(.(0, Xs)) → U3(Xs, p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(X, Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out(.(X, []))
U1(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U2(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(.(s(s(s(s(Y)))), Xs))) → p_out(.(s(s(X)), .(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(.(0, Xs))

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

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(.(0, Xs)) → U3(Xs, p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(X, Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out(.(X, []))
U1(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U2(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(.(s(s(s(s(Y)))), Xs))) → p_out(.(s(s(X)), .(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(.(0, Xs))

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


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(.(0, Xs)) → U31(Xs, p_in(Xs))
P_IN(.(0, Xs)) → P_IN(Xs)
P_IN(.(s(s(X)), .(Y, Xs))) → U11(X, Y, Xs, p_in(.(X, .(Y, Xs))))
P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
U11(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U21(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U11(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → P_IN(.(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in(.(0, Xs)) → U3(Xs, p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(X, Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out(.(X, []))
U1(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U2(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(.(s(s(s(s(Y)))), Xs))) → p_out(.(s(s(X)), .(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(.(0, Xs))

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

P_IN(.(0, Xs)) → U31(Xs, p_in(Xs))
P_IN(.(0, Xs)) → P_IN(Xs)
P_IN(.(s(s(X)), .(Y, Xs))) → U11(X, Y, Xs, p_in(.(X, .(Y, Xs))))
P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
U11(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U21(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U11(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → P_IN(.(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in(.(0, Xs)) → U3(Xs, p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(X, Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out(.(X, []))
U1(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U2(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(.(s(s(s(s(Y)))), Xs))) → p_out(.(s(s(X)), .(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(.(0, Xs))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
.(x1, x2)  =  .(x1, x2)
0  =  0
U3(x1, x2)  =  U3(x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
[]  =  []
p_out(x1)  =  p_out
U2(x1, x2, x3, x4)  =  U2(x4)
P_IN(x1)  =  P_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U31(x1, x2)  =  U31(x2)
U11(x1, x2, x3, x4)  =  U11(x2, x3, 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
              ↳ PiDPToQDPProof

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

P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
P_IN(.(s(s(X)), .(Y, Xs))) → U11(X, Y, Xs, p_in(.(X, .(Y, Xs))))
P_IN(.(0, Xs)) → P_IN(Xs)
U11(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → P_IN(.(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in(.(0, Xs)) → U3(Xs, p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(X, Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out(.(X, []))
U1(X, Y, Xs, p_out(.(X, .(Y, Xs)))) → U2(X, Y, Xs, p_in(.(s(s(s(s(Y)))), Xs)))
U2(X, Y, Xs, p_out(.(s(s(s(s(Y)))), Xs))) → p_out(.(s(s(X)), .(Y, Xs)))
U3(Xs, p_out(Xs)) → p_out(.(0, Xs))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
.(x1, x2)  =  .(x1, x2)
0  =  0
U3(x1, x2)  =  U3(x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x2, x3, x4)
[]  =  []
p_out(x1)  =  p_out
U2(x1, x2, x3, x4)  =  U2(x4)
P_IN(x1)  =  P_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x2, x3, 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
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof

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

P_IN(.(s(s(X)), .(Y, Xs))) → U11(Y, Xs, p_in(.(X, .(Y, Xs))))
P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
U11(Y, Xs, p_out) → P_IN(.(s(s(s(s(Y)))), Xs))
P_IN(.(0, Xs)) → P_IN(Xs)

The TRS R consists of the following rules:

p_in(.(0, Xs)) → U3(p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out
U1(Y, Xs, p_out) → U2(p_in(.(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
U3(p_out) → p_out

The set Q consists of the following terms:

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


P_IN(.(s(s(X)), .(Y, Xs))) → U11(Y, Xs, p_in(.(X, .(Y, Xs))))
P_IN(.(0, Xs)) → P_IN(Xs)
The remaining pairs can at least be oriented weakly.

P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
U11(Y, Xs, p_out) → P_IN(.(s(s(s(s(Y)))), Xs))
Used ordering: Polynomial interpretation [25]:

POL(.(x1, x2)) = 1 + x1 + x2   
POL(0) = 1   
POL(P_IN(x1)) = x1   
POL(U1(x1, x2, x3)) = 0   
POL(U11(x1, x2, x3)) = 1 + x1 + x2   
POL(U2(x1)) = 0   
POL(U3(x1)) = 0   
POL([]) = 0   
POL(p_in(x1)) = 1 + x1   
POL(p_out) = 1   
POL(s(x1)) = x1   

The following usable rules [17] were oriented: none



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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

P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
U11(Y, Xs, p_out) → P_IN(.(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in(.(0, Xs)) → U3(p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out
U1(Y, Xs, p_out) → U2(p_in(.(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
U3(p_out) → p_out

The set Q consists of the following terms:

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(x0)

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

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

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

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

The TRS R consists of the following rules:

p_in(.(0, Xs)) → U3(p_in(Xs))
p_in(.(s(s(X)), .(Y, Xs))) → U1(Y, Xs, p_in(.(X, .(Y, Xs))))
p_in(.(X, [])) → p_out
U1(Y, Xs, p_out) → U2(p_in(.(s(s(s(s(Y)))), Xs)))
U2(p_out) → p_out
U3(p_out) → p_out

The set Q consists of the following terms:

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof

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

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

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

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p_in(x0)
U1(x0, x1, x2)
U2(x0)
U3(x0)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ UsableRulesReductionPairsProof

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

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

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

P_IN(.(s(s(X)), .(Y, Xs))) → P_IN(.(X, .(Y, Xs)))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2·x1 + x2   
POL(P_IN(x1)) = 2·x1   
POL(s(x1)) = 2·x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ UsableRulesReductionPairsProof
                                    ↳ QDP
                                      ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.