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



PROLOG
  ↳ PrologToPiTRSProof

p2(X, g1(X)).
p2(X, f1(Y)) :- p2(X, g1(Y)).


With regard to the inferred argument filtering the predicates were used in the following modes:
p2: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


p_2_in_ga2(X, g_11(X)) -> p_2_out_ga2(X, g_11(X))
p_2_in_ga2(X, f_11(Y)) -> if_p_2_in_1_ga3(X, Y, p_2_in_ga2(X, g_11(Y)))
if_p_2_in_1_ga3(X, Y, p_2_out_ga2(X, g_11(Y))) -> p_2_out_ga2(X, f_11(Y))

The argument filtering Pi contains the following mapping:
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
g_11(x1)  =  g_11(x1)
f_11(x1)  =  f_11(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_p_2_in_1_ga3(x1, x2, x3)  =  if_p_2_in_1_ga1(x3)

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:

p_2_in_ga2(X, g_11(X)) -> p_2_out_ga2(X, g_11(X))
p_2_in_ga2(X, f_11(Y)) -> if_p_2_in_1_ga3(X, Y, p_2_in_ga2(X, g_11(Y)))
if_p_2_in_1_ga3(X, Y, p_2_out_ga2(X, g_11(Y))) -> p_2_out_ga2(X, f_11(Y))

The argument filtering Pi contains the following mapping:
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
g_11(x1)  =  g_11(x1)
f_11(x1)  =  f_11(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_p_2_in_1_ga3(x1, x2, x3)  =  if_p_2_in_1_ga1(x3)


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

P_2_IN_GA2(X, f_11(Y)) -> IF_P_2_IN_1_GA3(X, Y, p_2_in_ga2(X, g_11(Y)))
P_2_IN_GA2(X, f_11(Y)) -> P_2_IN_GA2(X, g_11(Y))

The TRS R consists of the following rules:

p_2_in_ga2(X, g_11(X)) -> p_2_out_ga2(X, g_11(X))
p_2_in_ga2(X, f_11(Y)) -> if_p_2_in_1_ga3(X, Y, p_2_in_ga2(X, g_11(Y)))
if_p_2_in_1_ga3(X, Y, p_2_out_ga2(X, g_11(Y))) -> p_2_out_ga2(X, f_11(Y))

The argument filtering Pi contains the following mapping:
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
g_11(x1)  =  g_11(x1)
f_11(x1)  =  f_11(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_p_2_in_1_ga3(x1, x2, x3)  =  if_p_2_in_1_ga1(x3)
IF_P_2_IN_1_GA3(x1, x2, x3)  =  IF_P_2_IN_1_GA1(x3)
P_2_IN_GA2(x1, x2)  =  P_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:

P_2_IN_GA2(X, f_11(Y)) -> IF_P_2_IN_1_GA3(X, Y, p_2_in_ga2(X, g_11(Y)))
P_2_IN_GA2(X, f_11(Y)) -> P_2_IN_GA2(X, g_11(Y))

The TRS R consists of the following rules:

p_2_in_ga2(X, g_11(X)) -> p_2_out_ga2(X, g_11(X))
p_2_in_ga2(X, f_11(Y)) -> if_p_2_in_1_ga3(X, Y, p_2_in_ga2(X, g_11(Y)))
if_p_2_in_1_ga3(X, Y, p_2_out_ga2(X, g_11(Y))) -> p_2_out_ga2(X, f_11(Y))

The argument filtering Pi contains the following mapping:
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
g_11(x1)  =  g_11(x1)
f_11(x1)  =  f_11(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_p_2_in_1_ga3(x1, x2, x3)  =  if_p_2_in_1_ga1(x3)
IF_P_2_IN_1_GA3(x1, x2, x3)  =  IF_P_2_IN_1_GA1(x3)
P_2_IN_GA2(x1, x2)  =  P_2_IN_GA1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.