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



↳ PROLOG
  ↳ PrologToPiTRSProof

sum3(X, s1(Y), s1(Z)) :- sum3(X, Y, Z).
sum3(X, 00, X).


With regard to the inferred argument filtering the predicates were used in the following modes:
sum3: (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


sum_3_in_aag3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aag4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
sum_3_in_aag3(X, 0_0, X) -> sum_3_out_aag3(X, 0_0, X)
if_sum_3_in_1_aag4(X, Y, Z, sum_3_out_aag3(X, Y, Z)) -> sum_3_out_aag3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aag3(x1, x2, x3)  =  sum_3_in_aag1(x3)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_sum_3_in_1_aag4(x1, x2, x3, x4)  =  if_sum_3_in_1_aag1(x4)
sum_3_out_aag3(x1, x2, x3)  =  sum_3_out_aag2(x1, 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:

sum_3_in_aag3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aag4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
sum_3_in_aag3(X, 0_0, X) -> sum_3_out_aag3(X, 0_0, X)
if_sum_3_in_1_aag4(X, Y, Z, sum_3_out_aag3(X, Y, Z)) -> sum_3_out_aag3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aag3(x1, x2, x3)  =  sum_3_in_aag1(x3)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_sum_3_in_1_aag4(x1, x2, x3, x4)  =  if_sum_3_in_1_aag1(x4)
sum_3_out_aag3(x1, x2, x3)  =  sum_3_out_aag2(x1, x2)


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

SUM_3_IN_AAG3(X, s_11(Y), s_11(Z)) -> IF_SUM_3_IN_1_AAG4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
SUM_3_IN_AAG3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AAG3(X, Y, Z)

The TRS R consists of the following rules:

sum_3_in_aag3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aag4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
sum_3_in_aag3(X, 0_0, X) -> sum_3_out_aag3(X, 0_0, X)
if_sum_3_in_1_aag4(X, Y, Z, sum_3_out_aag3(X, Y, Z)) -> sum_3_out_aag3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aag3(x1, x2, x3)  =  sum_3_in_aag1(x3)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_sum_3_in_1_aag4(x1, x2, x3, x4)  =  if_sum_3_in_1_aag1(x4)
sum_3_out_aag3(x1, x2, x3)  =  sum_3_out_aag2(x1, x2)
IF_SUM_3_IN_1_AAG4(x1, x2, x3, x4)  =  IF_SUM_3_IN_1_AAG1(x4)
SUM_3_IN_AAG3(x1, x2, x3)  =  SUM_3_IN_AAG1(x3)

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:

SUM_3_IN_AAG3(X, s_11(Y), s_11(Z)) -> IF_SUM_3_IN_1_AAG4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
SUM_3_IN_AAG3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AAG3(X, Y, Z)

The TRS R consists of the following rules:

sum_3_in_aag3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aag4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
sum_3_in_aag3(X, 0_0, X) -> sum_3_out_aag3(X, 0_0, X)
if_sum_3_in_1_aag4(X, Y, Z, sum_3_out_aag3(X, Y, Z)) -> sum_3_out_aag3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aag3(x1, x2, x3)  =  sum_3_in_aag1(x3)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_sum_3_in_1_aag4(x1, x2, x3, x4)  =  if_sum_3_in_1_aag1(x4)
sum_3_out_aag3(x1, x2, x3)  =  sum_3_out_aag2(x1, x2)
IF_SUM_3_IN_1_AAG4(x1, x2, x3, x4)  =  IF_SUM_3_IN_1_AAG1(x4)
SUM_3_IN_AAG3(x1, x2, x3)  =  SUM_3_IN_AAG1(x3)

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:

SUM_3_IN_AAG3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AAG3(X, Y, Z)

The TRS R consists of the following rules:

sum_3_in_aag3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aag4(X, Y, Z, sum_3_in_aag3(X, Y, Z))
sum_3_in_aag3(X, 0_0, X) -> sum_3_out_aag3(X, 0_0, X)
if_sum_3_in_1_aag4(X, Y, Z, sum_3_out_aag3(X, Y, Z)) -> sum_3_out_aag3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aag3(x1, x2, x3)  =  sum_3_in_aag1(x3)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_sum_3_in_1_aag4(x1, x2, x3, x4)  =  if_sum_3_in_1_aag1(x4)
sum_3_out_aag3(x1, x2, x3)  =  sum_3_out_aag2(x1, x2)
SUM_3_IN_AAG3(x1, x2, x3)  =  SUM_3_IN_AAG1(x3)

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:

SUM_3_IN_AAG3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AAG3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
SUM_3_IN_AAG3(x1, x2, x3)  =  SUM_3_IN_AAG1(x3)

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:

SUM_3_IN_AAG1(s_11(Z)) -> SUM_3_IN_AAG1(Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SUM_3_IN_AAG1}.
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: