Left Termination of the query pattern len1(b,f) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ PrologToPiTRSProof

len12({}0, 00).
len12(.2(underscore, Ts), N) :- len12(Ts, M), eq2(N, s1(M)).
eq2(X, X).


With regard to the inferred argument filtering the predicates were used in the following modes:
len12: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


len1_2_in_ga2([]_0, 0_0) -> len1_2_out_ga2([]_0, 0_0)
len1_2_in_ga2(._22(underscore, Ts), N) -> if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
eq_2_in_ag2(X, X) -> eq_2_out_ag2(X, X)
if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_out_ag2(N, s_11(M))) -> len1_2_out_ga2(._22(underscore, Ts), N)

The argument filtering Pi contains the following mapping:
len1_2_in_ga2(x1, x2)  =  len1_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len1_2_out_ga2(x1, x2)  =  len1_2_out_ga1(x2)
if_len1_2_in_1_ga4(x1, x2, x3, x4)  =  if_len1_2_in_1_ga1(x4)
if_len1_2_in_2_ga5(x1, x2, x3, x4, x5)  =  if_len1_2_in_2_ga1(x5)
eq_2_in_ag2(x1, x2)  =  eq_2_in_ag1(x2)
eq_2_out_ag2(x1, x2)  =  eq_2_out_ag1(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_2_in_ga2([]_0, 0_0) -> len1_2_out_ga2([]_0, 0_0)
len1_2_in_ga2(._22(underscore, Ts), N) -> if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
eq_2_in_ag2(X, X) -> eq_2_out_ag2(X, X)
if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_out_ag2(N, s_11(M))) -> len1_2_out_ga2(._22(underscore, Ts), N)

The argument filtering Pi contains the following mapping:
len1_2_in_ga2(x1, x2)  =  len1_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len1_2_out_ga2(x1, x2)  =  len1_2_out_ga1(x2)
if_len1_2_in_1_ga4(x1, x2, x3, x4)  =  if_len1_2_in_1_ga1(x4)
if_len1_2_in_2_ga5(x1, x2, x3, x4, x5)  =  if_len1_2_in_2_ga1(x5)
eq_2_in_ag2(x1, x2)  =  eq_2_in_ag1(x2)
eq_2_out_ag2(x1, x2)  =  eq_2_out_ag1(x1)


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

LEN1_2_IN_GA2(._22(underscore, Ts), N) -> IF_LEN1_2_IN_1_GA4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
LEN1_2_IN_GA2(._22(underscore, Ts), N) -> LEN1_2_IN_GA2(Ts, M)
IF_LEN1_2_IN_1_GA4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> IF_LEN1_2_IN_2_GA5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
IF_LEN1_2_IN_1_GA4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> EQ_2_IN_AG2(N, s_11(M))

The TRS R consists of the following rules:

len1_2_in_ga2([]_0, 0_0) -> len1_2_out_ga2([]_0, 0_0)
len1_2_in_ga2(._22(underscore, Ts), N) -> if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
eq_2_in_ag2(X, X) -> eq_2_out_ag2(X, X)
if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_out_ag2(N, s_11(M))) -> len1_2_out_ga2(._22(underscore, Ts), N)

The argument filtering Pi contains the following mapping:
len1_2_in_ga2(x1, x2)  =  len1_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len1_2_out_ga2(x1, x2)  =  len1_2_out_ga1(x2)
if_len1_2_in_1_ga4(x1, x2, x3, x4)  =  if_len1_2_in_1_ga1(x4)
if_len1_2_in_2_ga5(x1, x2, x3, x4, x5)  =  if_len1_2_in_2_ga1(x5)
eq_2_in_ag2(x1, x2)  =  eq_2_in_ag1(x2)
eq_2_out_ag2(x1, x2)  =  eq_2_out_ag1(x1)
EQ_2_IN_AG2(x1, x2)  =  EQ_2_IN_AG1(x2)
IF_LEN1_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_LEN1_2_IN_1_GA1(x4)
IF_LEN1_2_IN_2_GA5(x1, x2, x3, x4, x5)  =  IF_LEN1_2_IN_2_GA1(x5)
LEN1_2_IN_GA2(x1, x2)  =  LEN1_2_IN_GA1(x1)

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_2_IN_GA2(._22(underscore, Ts), N) -> IF_LEN1_2_IN_1_GA4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
LEN1_2_IN_GA2(._22(underscore, Ts), N) -> LEN1_2_IN_GA2(Ts, M)
IF_LEN1_2_IN_1_GA4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> IF_LEN1_2_IN_2_GA5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
IF_LEN1_2_IN_1_GA4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> EQ_2_IN_AG2(N, s_11(M))

The TRS R consists of the following rules:

len1_2_in_ga2([]_0, 0_0) -> len1_2_out_ga2([]_0, 0_0)
len1_2_in_ga2(._22(underscore, Ts), N) -> if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
eq_2_in_ag2(X, X) -> eq_2_out_ag2(X, X)
if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_out_ag2(N, s_11(M))) -> len1_2_out_ga2(._22(underscore, Ts), N)

The argument filtering Pi contains the following mapping:
len1_2_in_ga2(x1, x2)  =  len1_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len1_2_out_ga2(x1, x2)  =  len1_2_out_ga1(x2)
if_len1_2_in_1_ga4(x1, x2, x3, x4)  =  if_len1_2_in_1_ga1(x4)
if_len1_2_in_2_ga5(x1, x2, x3, x4, x5)  =  if_len1_2_in_2_ga1(x5)
eq_2_in_ag2(x1, x2)  =  eq_2_in_ag1(x2)
eq_2_out_ag2(x1, x2)  =  eq_2_out_ag1(x1)
EQ_2_IN_AG2(x1, x2)  =  EQ_2_IN_AG1(x2)
IF_LEN1_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_LEN1_2_IN_1_GA1(x4)
IF_LEN1_2_IN_2_GA5(x1, x2, x3, x4, x5)  =  IF_LEN1_2_IN_2_GA1(x5)
LEN1_2_IN_GA2(x1, x2)  =  LEN1_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph 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_2_IN_GA2(._22(underscore, Ts), N) -> LEN1_2_IN_GA2(Ts, M)

The TRS R consists of the following rules:

len1_2_in_ga2([]_0, 0_0) -> len1_2_out_ga2([]_0, 0_0)
len1_2_in_ga2(._22(underscore, Ts), N) -> if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_in_ga2(Ts, M))
if_len1_2_in_1_ga4(underscore, Ts, N, len1_2_out_ga2(Ts, M)) -> if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_in_ag2(N, s_11(M)))
eq_2_in_ag2(X, X) -> eq_2_out_ag2(X, X)
if_len1_2_in_2_ga5(underscore, Ts, N, M, eq_2_out_ag2(N, s_11(M))) -> len1_2_out_ga2(._22(underscore, Ts), N)

The argument filtering Pi contains the following mapping:
len1_2_in_ga2(x1, x2)  =  len1_2_in_ga1(x1)
[]_0  =  []_0
0_0  =  0_0
._22(x1, x2)  =  ._22(x1, x2)
s_11(x1)  =  s_11(x1)
len1_2_out_ga2(x1, x2)  =  len1_2_out_ga1(x2)
if_len1_2_in_1_ga4(x1, x2, x3, x4)  =  if_len1_2_in_1_ga1(x4)
if_len1_2_in_2_ga5(x1, x2, x3, x4, x5)  =  if_len1_2_in_2_ga1(x5)
eq_2_in_ag2(x1, x2)  =  eq_2_in_ag1(x2)
eq_2_out_ag2(x1, x2)  =  eq_2_out_ag1(x1)
LEN1_2_IN_GA2(x1, x2)  =  LEN1_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:

LEN1_2_IN_GA2(._22(underscore, Ts), N) -> LEN1_2_IN_GA2(Ts, M)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
LEN1_2_IN_GA2(x1, x2)  =  LEN1_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:

LEN1_2_IN_GA1(._22(underscore, Ts)) -> LEN1_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 {LEN1_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: