Left Termination of the query pattern len(b,f) w.r.t. the given Prolog program could successfully be proven:



PROLOG
  ↳ PrologToPiTRSProof

len2({}0, 00).
len2(.2(underscore, Ts), s1(N)) :- len2(Ts, N).


With regard to the inferred argument filtering the predicates were used in the following modes:
len2: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


len_2_in_ga2([]_0, 0_0) -> len_2_out_ga2([]_0, 0_0)
len_2_in_ga2(._22(underscore, Ts), s_11(N)) -> if_len_2_in_1_ga4(underscore, Ts, N, len_2_in_ga2(Ts, N))
if_len_2_in_1_ga4(underscore, Ts, N, len_2_out_ga2(Ts, N)) -> len_2_out_ga2(._22(underscore, Ts), s_11(N))

The argument filtering Pi contains the following mapping:
len_2_in_ga2(x1, x2)  =  len_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len_2_out_ga2(x1, x2)  =  len_2_out_ga1(x2)
if_len_2_in_1_ga4(x1, x2, x3, x4)  =  if_len_2_in_1_ga1(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:

len_2_in_ga2([]_0, 0_0) -> len_2_out_ga2([]_0, 0_0)
len_2_in_ga2(._22(underscore, Ts), s_11(N)) -> if_len_2_in_1_ga4(underscore, Ts, N, len_2_in_ga2(Ts, N))
if_len_2_in_1_ga4(underscore, Ts, N, len_2_out_ga2(Ts, N)) -> len_2_out_ga2(._22(underscore, Ts), s_11(N))

The argument filtering Pi contains the following mapping:
len_2_in_ga2(x1, x2)  =  len_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len_2_out_ga2(x1, x2)  =  len_2_out_ga1(x2)
if_len_2_in_1_ga4(x1, x2, x3, x4)  =  if_len_2_in_1_ga1(x4)


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

LEN_2_IN_GA2(._22(underscore, Ts), s_11(N)) -> IF_LEN_2_IN_1_GA4(underscore, Ts, N, len_2_in_ga2(Ts, N))
LEN_2_IN_GA2(._22(underscore, Ts), s_11(N)) -> LEN_2_IN_GA2(Ts, N)

The TRS R consists of the following rules:

len_2_in_ga2([]_0, 0_0) -> len_2_out_ga2([]_0, 0_0)
len_2_in_ga2(._22(underscore, Ts), s_11(N)) -> if_len_2_in_1_ga4(underscore, Ts, N, len_2_in_ga2(Ts, N))
if_len_2_in_1_ga4(underscore, Ts, N, len_2_out_ga2(Ts, N)) -> len_2_out_ga2(._22(underscore, Ts), s_11(N))

The argument filtering Pi contains the following mapping:
len_2_in_ga2(x1, x2)  =  len_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len_2_out_ga2(x1, x2)  =  len_2_out_ga1(x2)
if_len_2_in_1_ga4(x1, x2, x3, x4)  =  if_len_2_in_1_ga1(x4)
LEN_2_IN_GA2(x1, x2)  =  LEN_2_IN_GA1(x1)
IF_LEN_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_LEN_2_IN_1_GA1(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_2_IN_GA2(._22(underscore, Ts), s_11(N)) -> IF_LEN_2_IN_1_GA4(underscore, Ts, N, len_2_in_ga2(Ts, N))
LEN_2_IN_GA2(._22(underscore, Ts), s_11(N)) -> LEN_2_IN_GA2(Ts, N)

The TRS R consists of the following rules:

len_2_in_ga2([]_0, 0_0) -> len_2_out_ga2([]_0, 0_0)
len_2_in_ga2(._22(underscore, Ts), s_11(N)) -> if_len_2_in_1_ga4(underscore, Ts, N, len_2_in_ga2(Ts, N))
if_len_2_in_1_ga4(underscore, Ts, N, len_2_out_ga2(Ts, N)) -> len_2_out_ga2(._22(underscore, Ts), s_11(N))

The argument filtering Pi contains the following mapping:
len_2_in_ga2(x1, x2)  =  len_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len_2_out_ga2(x1, x2)  =  len_2_out_ga1(x2)
if_len_2_in_1_ga4(x1, x2, x3, x4)  =  if_len_2_in_1_ga1(x4)
LEN_2_IN_GA2(x1, x2)  =  LEN_2_IN_GA1(x1)
IF_LEN_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_LEN_2_IN_1_GA1(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph 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_2_IN_GA2(._22(underscore, Ts), s_11(N)) -> LEN_2_IN_GA2(Ts, N)

The TRS R consists of the following rules:

len_2_in_ga2([]_0, 0_0) -> len_2_out_ga2([]_0, 0_0)
len_2_in_ga2(._22(underscore, Ts), s_11(N)) -> if_len_2_in_1_ga4(underscore, Ts, N, len_2_in_ga2(Ts, N))
if_len_2_in_1_ga4(underscore, Ts, N, len_2_out_ga2(Ts, N)) -> len_2_out_ga2(._22(underscore, Ts), s_11(N))

The argument filtering Pi contains the following mapping:
len_2_in_ga2(x1, x2)  =  len_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len_2_out_ga2(x1, x2)  =  len_2_out_ga1(x2)
if_len_2_in_1_ga4(x1, x2, x3, x4)  =  if_len_2_in_1_ga1(x4)
LEN_2_IN_GA2(x1, x2)  =  LEN_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_2_IN_GA2(._22(underscore, Ts), s_11(N)) -> LEN_2_IN_GA2(Ts, N)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
LEN_2_IN_GA2(x1, x2)  =  LEN_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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_2_IN_GA1(._22(underscore, Ts)) -> LEN_2_IN_GA1(Ts)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LEN_2_IN_GA1}.
By using the subterm criterion together with the size-change analysis 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: