Left Termination of the query pattern len_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

len([], 0).
len(.(X, Ts), s(N)) :- len(Ts, N).

Queries:

len(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:

len_in(.(X, Ts), s(N)) → U1(X, Ts, N, len_in(Ts, N))
len_in([], 0) → len_out([], 0)
U1(X, Ts, N, len_out(Ts, N)) → len_out(.(X, Ts), s(N))

The argument filtering Pi contains the following mapping:
len_in(x1, x2)  =  len_in(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len_out(x1, x2)  =  len_out(x2)

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:

len_in(.(X, Ts), s(N)) → U1(X, Ts, N, len_in(Ts, N))
len_in([], 0) → len_out([], 0)
U1(X, Ts, N, len_out(Ts, N)) → len_out(.(X, Ts), s(N))

The argument filtering Pi contains the following mapping:
len_in(x1, x2)  =  len_in(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len_out(x1, x2)  =  len_out(x2)


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

LEN_IN(.(X, Ts), s(N)) → U11(X, Ts, N, len_in(Ts, N))
LEN_IN(.(X, Ts), s(N)) → LEN_IN(Ts, N)

The TRS R consists of the following rules:

len_in(.(X, Ts), s(N)) → U1(X, Ts, N, len_in(Ts, N))
len_in([], 0) → len_out([], 0)
U1(X, Ts, N, len_out(Ts, N)) → len_out(.(X, Ts), s(N))

The argument filtering Pi contains the following mapping:
len_in(x1, x2)  =  len_in(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len_out(x1, x2)  =  len_out(x2)
LEN_IN(x1, x2)  =  LEN_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:

LEN_IN(.(X, Ts), s(N)) → U11(X, Ts, N, len_in(Ts, N))
LEN_IN(.(X, Ts), s(N)) → LEN_IN(Ts, N)

The TRS R consists of the following rules:

len_in(.(X, Ts), s(N)) → U1(X, Ts, N, len_in(Ts, N))
len_in([], 0) → len_out([], 0)
U1(X, Ts, N, len_out(Ts, N)) → len_out(.(X, Ts), s(N))

The argument filtering Pi contains the following mapping:
len_in(x1, x2)  =  len_in(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len_out(x1, x2)  =  len_out(x2)
LEN_IN(x1, x2)  =  LEN_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 1 less node.

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

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

LEN_IN(.(X, Ts), s(N)) → LEN_IN(Ts, N)

The TRS R consists of the following rules:

len_in(.(X, Ts), s(N)) → U1(X, Ts, N, len_in(Ts, N))
len_in([], 0) → len_out([], 0)
U1(X, Ts, N, len_out(Ts, N)) → len_out(.(X, Ts), s(N))

The argument filtering Pi contains the following mapping:
len_in(x1, x2)  =  len_in(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len_out(x1, x2)  =  len_out(x2)
LEN_IN(x1, x2)  =  LEN_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:

LEN_IN(.(X, Ts), s(N)) → LEN_IN(Ts, N)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
LEN_IN(x1, x2)  =  LEN_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

LEN_IN(.(X, Ts)) → LEN_IN(Ts)

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: