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



↳ PROLOG
  ↳ PrologToPiTRSProof

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


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


sum_3_in_aga3(X, 0_0, X) -> sum_3_out_aga3(X, 0_0, X)
sum_3_in_aga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aga4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
if_sum_3_in_1_aga4(X, Y, Z, sum_3_out_aga3(X, Y, Z)) -> sum_3_out_aga3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aga3(x1, x2, x3)  =  sum_3_in_aga1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
sum_3_out_aga3(x1, x2, x3)  =  sum_3_out_aga
if_sum_3_in_1_aga4(x1, x2, x3, x4)  =  if_sum_3_in_1_aga1(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:

sum_3_in_aga3(X, 0_0, X) -> sum_3_out_aga3(X, 0_0, X)
sum_3_in_aga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aga4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
if_sum_3_in_1_aga4(X, Y, Z, sum_3_out_aga3(X, Y, Z)) -> sum_3_out_aga3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aga3(x1, x2, x3)  =  sum_3_in_aga1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
sum_3_out_aga3(x1, x2, x3)  =  sum_3_out_aga
if_sum_3_in_1_aga4(x1, x2, x3, x4)  =  if_sum_3_in_1_aga1(x4)


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

SUM_3_IN_AGA3(X, s_11(Y), s_11(Z)) -> IF_SUM_3_IN_1_AGA4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
SUM_3_IN_AGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AGA3(X, Y, Z)

The TRS R consists of the following rules:

sum_3_in_aga3(X, 0_0, X) -> sum_3_out_aga3(X, 0_0, X)
sum_3_in_aga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aga4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
if_sum_3_in_1_aga4(X, Y, Z, sum_3_out_aga3(X, Y, Z)) -> sum_3_out_aga3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aga3(x1, x2, x3)  =  sum_3_in_aga1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
sum_3_out_aga3(x1, x2, x3)  =  sum_3_out_aga
if_sum_3_in_1_aga4(x1, x2, x3, x4)  =  if_sum_3_in_1_aga1(x4)
SUM_3_IN_AGA3(x1, x2, x3)  =  SUM_3_IN_AGA1(x2)
IF_SUM_3_IN_1_AGA4(x1, x2, x3, x4)  =  IF_SUM_3_IN_1_AGA1(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:

SUM_3_IN_AGA3(X, s_11(Y), s_11(Z)) -> IF_SUM_3_IN_1_AGA4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
SUM_3_IN_AGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AGA3(X, Y, Z)

The TRS R consists of the following rules:

sum_3_in_aga3(X, 0_0, X) -> sum_3_out_aga3(X, 0_0, X)
sum_3_in_aga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aga4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
if_sum_3_in_1_aga4(X, Y, Z, sum_3_out_aga3(X, Y, Z)) -> sum_3_out_aga3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aga3(x1, x2, x3)  =  sum_3_in_aga1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
sum_3_out_aga3(x1, x2, x3)  =  sum_3_out_aga
if_sum_3_in_1_aga4(x1, x2, x3, x4)  =  if_sum_3_in_1_aga1(x4)
SUM_3_IN_AGA3(x1, x2, x3)  =  SUM_3_IN_AGA1(x2)
IF_SUM_3_IN_1_AGA4(x1, x2, x3, x4)  =  IF_SUM_3_IN_1_AGA1(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:

SUM_3_IN_AGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AGA3(X, Y, Z)

The TRS R consists of the following rules:

sum_3_in_aga3(X, 0_0, X) -> sum_3_out_aga3(X, 0_0, X)
sum_3_in_aga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_aga4(X, Y, Z, sum_3_in_aga3(X, Y, Z))
if_sum_3_in_1_aga4(X, Y, Z, sum_3_out_aga3(X, Y, Z)) -> sum_3_out_aga3(X, s_11(Y), s_11(Z))

The argument filtering Pi contains the following mapping:
sum_3_in_aga3(x1, x2, x3)  =  sum_3_in_aga1(x2)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
sum_3_out_aga3(x1, x2, x3)  =  sum_3_out_aga
if_sum_3_in_1_aga4(x1, x2, x3, x4)  =  if_sum_3_in_1_aga1(x4)
SUM_3_IN_AGA3(x1, x2, x3)  =  SUM_3_IN_AGA1(x2)

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_AGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_AGA3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
SUM_3_IN_AGA3(x1, x2, x3)  =  SUM_3_IN_AGA1(x2)

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_AGA1(s_11(Y)) -> SUM_3_IN_AGA1(Y)

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_AGA1}.
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: