Left Termination of the query pattern len1_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

len1([], 0).
len1(.(X, Ts), N) :- ','(len1(Ts, M), eq(N, s(M))).
eq(X, X).

Queries:

len1(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:

len1_in(.(X, Ts), N) → U1(X, Ts, N, len1_in(Ts, M))
len1_in([], 0) → len1_out([], 0)
U1(X, Ts, N, len1_out(Ts, M)) → U2(X, Ts, N, M, eq_in(N, s(M)))
eq_in(X, X) → eq_out(X, X)
U2(X, Ts, N, M, eq_out(N, s(M))) → len1_out(.(X, Ts), N)

The argument filtering Pi contains the following mapping:
len1_in(x1, x2)  =  len1_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len1_out(x1, x2)  =  len1_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
eq_in(x1, x2)  =  eq_in(x2)
s(x1)  =  s(x1)
eq_out(x1, x2)  =  eq_out(x1)

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:

len1_in(.(X, Ts), N) → U1(X, Ts, N, len1_in(Ts, M))
len1_in([], 0) → len1_out([], 0)
U1(X, Ts, N, len1_out(Ts, M)) → U2(X, Ts, N, M, eq_in(N, s(M)))
eq_in(X, X) → eq_out(X, X)
U2(X, Ts, N, M, eq_out(N, s(M))) → len1_out(.(X, Ts), N)

The argument filtering Pi contains the following mapping:
len1_in(x1, x2)  =  len1_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len1_out(x1, x2)  =  len1_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
eq_in(x1, x2)  =  eq_in(x2)
s(x1)  =  s(x1)
eq_out(x1, x2)  =  eq_out(x1)


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

LEN1_IN(.(X, Ts), N) → U11(X, Ts, N, len1_in(Ts, M))
LEN1_IN(.(X, Ts), N) → LEN1_IN(Ts, M)
U11(X, Ts, N, len1_out(Ts, M)) → U21(X, Ts, N, M, eq_in(N, s(M)))
U11(X, Ts, N, len1_out(Ts, M)) → EQ_IN(N, s(M))

The TRS R consists of the following rules:

len1_in(.(X, Ts), N) → U1(X, Ts, N, len1_in(Ts, M))
len1_in([], 0) → len1_out([], 0)
U1(X, Ts, N, len1_out(Ts, M)) → U2(X, Ts, N, M, eq_in(N, s(M)))
eq_in(X, X) → eq_out(X, X)
U2(X, Ts, N, M, eq_out(N, s(M))) → len1_out(.(X, Ts), N)

The argument filtering Pi contains the following mapping:
len1_in(x1, x2)  =  len1_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len1_out(x1, x2)  =  len1_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
eq_in(x1, x2)  =  eq_in(x2)
s(x1)  =  s(x1)
eq_out(x1, x2)  =  eq_out(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
LEN1_IN(x1, x2)  =  LEN1_IN(x1)
EQ_IN(x1, x2)  =  EQ_IN(x2)
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:

LEN1_IN(.(X, Ts), N) → U11(X, Ts, N, len1_in(Ts, M))
LEN1_IN(.(X, Ts), N) → LEN1_IN(Ts, M)
U11(X, Ts, N, len1_out(Ts, M)) → U21(X, Ts, N, M, eq_in(N, s(M)))
U11(X, Ts, N, len1_out(Ts, M)) → EQ_IN(N, s(M))

The TRS R consists of the following rules:

len1_in(.(X, Ts), N) → U1(X, Ts, N, len1_in(Ts, M))
len1_in([], 0) → len1_out([], 0)
U1(X, Ts, N, len1_out(Ts, M)) → U2(X, Ts, N, M, eq_in(N, s(M)))
eq_in(X, X) → eq_out(X, X)
U2(X, Ts, N, M, eq_out(N, s(M))) → len1_out(.(X, Ts), N)

The argument filtering Pi contains the following mapping:
len1_in(x1, x2)  =  len1_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len1_out(x1, x2)  =  len1_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
eq_in(x1, x2)  =  eq_in(x2)
s(x1)  =  s(x1)
eq_out(x1, x2)  =  eq_out(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
LEN1_IN(x1, x2)  =  LEN1_IN(x1)
EQ_IN(x1, x2)  =  EQ_IN(x2)
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 3 less nodes.

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

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

LEN1_IN(.(X, Ts), N) → LEN1_IN(Ts, M)

The TRS R consists of the following rules:

len1_in(.(X, Ts), N) → U1(X, Ts, N, len1_in(Ts, M))
len1_in([], 0) → len1_out([], 0)
U1(X, Ts, N, len1_out(Ts, M)) → U2(X, Ts, N, M, eq_in(N, s(M)))
eq_in(X, X) → eq_out(X, X)
U2(X, Ts, N, M, eq_out(N, s(M))) → len1_out(.(X, Ts), N)

The argument filtering Pi contains the following mapping:
len1_in(x1, x2)  =  len1_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
[]  =  []
0  =  0
len1_out(x1, x2)  =  len1_out(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
eq_in(x1, x2)  =  eq_in(x2)
s(x1)  =  s(x1)
eq_out(x1, x2)  =  eq_out(x1)
LEN1_IN(x1, x2)  =  LEN1_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:

LEN1_IN(.(X, Ts), N) → LEN1_IN(Ts, M)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
LEN1_IN(x1, x2)  =  LEN1_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:

LEN1_IN(.(X, Ts)) → LEN1_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: