Left Termination of the query pattern num_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

num(0).
num(s(X)) :- num(X).

Queries:

num(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:

num_in(s(X)) → U1(X, num_in(X))
num_in(0) → num_out(0)
U1(X, num_out(X)) → num_out(s(X))

The argument filtering Pi contains the following mapping:
num_in(x1)  =  num_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
0  =  0
num_out(x1)  =  num_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:

num_in(s(X)) → U1(X, num_in(X))
num_in(0) → num_out(0)
U1(X, num_out(X)) → num_out(s(X))

The argument filtering Pi contains the following mapping:
num_in(x1)  =  num_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
0  =  0
num_out(x1)  =  num_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:

NUM_IN(s(X)) → U11(X, num_in(X))
NUM_IN(s(X)) → NUM_IN(X)

The TRS R consists of the following rules:

num_in(s(X)) → U1(X, num_in(X))
num_in(0) → num_out(0)
U1(X, num_out(X)) → num_out(s(X))

The argument filtering Pi contains the following mapping:
num_in(x1)  =  num_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
0  =  0
num_out(x1)  =  num_out
NUM_IN(x1)  =  NUM_IN(x1)
U11(x1, x2)  =  U11(x2)

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:

NUM_IN(s(X)) → U11(X, num_in(X))
NUM_IN(s(X)) → NUM_IN(X)

The TRS R consists of the following rules:

num_in(s(X)) → U1(X, num_in(X))
num_in(0) → num_out(0)
U1(X, num_out(X)) → num_out(s(X))

The argument filtering Pi contains the following mapping:
num_in(x1)  =  num_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
0  =  0
num_out(x1)  =  num_out
NUM_IN(x1)  =  NUM_IN(x1)
U11(x1, x2)  =  U11(x2)

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

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

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

NUM_IN(s(X)) → NUM_IN(X)

The TRS R consists of the following rules:

num_in(s(X)) → U1(X, num_in(X))
num_in(0) → num_out(0)
U1(X, num_out(X)) → num_out(s(X))

The argument filtering Pi contains the following mapping:
num_in(x1)  =  num_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
0  =  0
num_out(x1)  =  num_out
NUM_IN(x1)  =  NUM_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

NUM_IN(s(X)) → NUM_IN(X)

R is empty.
Pi is empty.
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:

NUM_IN(s(X)) → NUM_IN(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: