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



↳ PROLOG
  ↳ PrologToPiTRSProof

plus3(00, Y, Y).
plus3(s1(X), Y, Z) :- plus3(X, s1(Y), Z).


With regard to the inferred argument filtering the predicates were used in the following modes:
plus3: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


plus_3_in_gaa3(0_0, Y, Y) -> plus_3_out_gaa3(0_0, Y, Y)
plus_3_in_gaa3(s_11(X), Y, Z) -> if_plus_3_in_1_gaa4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
if_plus_3_in_1_gaa4(X, Y, Z, plus_3_out_gaa3(X, s_11(Y), Z)) -> plus_3_out_gaa3(s_11(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
plus_3_out_gaa3(x1, x2, x3)  =  plus_3_out_gaa
if_plus_3_in_1_gaa4(x1, x2, x3, x4)  =  if_plus_3_in_1_gaa1(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:

plus_3_in_gaa3(0_0, Y, Y) -> plus_3_out_gaa3(0_0, Y, Y)
plus_3_in_gaa3(s_11(X), Y, Z) -> if_plus_3_in_1_gaa4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
if_plus_3_in_1_gaa4(X, Y, Z, plus_3_out_gaa3(X, s_11(Y), Z)) -> plus_3_out_gaa3(s_11(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
plus_3_out_gaa3(x1, x2, x3)  =  plus_3_out_gaa
if_plus_3_in_1_gaa4(x1, x2, x3, x4)  =  if_plus_3_in_1_gaa1(x4)


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

PLUS_3_IN_GAA3(s_11(X), Y, Z) -> IF_PLUS_3_IN_1_GAA4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
PLUS_3_IN_GAA3(s_11(X), Y, Z) -> PLUS_3_IN_GAA3(X, s_11(Y), Z)

The TRS R consists of the following rules:

plus_3_in_gaa3(0_0, Y, Y) -> plus_3_out_gaa3(0_0, Y, Y)
plus_3_in_gaa3(s_11(X), Y, Z) -> if_plus_3_in_1_gaa4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
if_plus_3_in_1_gaa4(X, Y, Z, plus_3_out_gaa3(X, s_11(Y), Z)) -> plus_3_out_gaa3(s_11(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
plus_3_out_gaa3(x1, x2, x3)  =  plus_3_out_gaa
if_plus_3_in_1_gaa4(x1, x2, x3, x4)  =  if_plus_3_in_1_gaa1(x4)
IF_PLUS_3_IN_1_GAA4(x1, x2, x3, x4)  =  IF_PLUS_3_IN_1_GAA1(x4)
PLUS_3_IN_GAA3(x1, x2, x3)  =  PLUS_3_IN_GAA1(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:

PLUS_3_IN_GAA3(s_11(X), Y, Z) -> IF_PLUS_3_IN_1_GAA4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
PLUS_3_IN_GAA3(s_11(X), Y, Z) -> PLUS_3_IN_GAA3(X, s_11(Y), Z)

The TRS R consists of the following rules:

plus_3_in_gaa3(0_0, Y, Y) -> plus_3_out_gaa3(0_0, Y, Y)
plus_3_in_gaa3(s_11(X), Y, Z) -> if_plus_3_in_1_gaa4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
if_plus_3_in_1_gaa4(X, Y, Z, plus_3_out_gaa3(X, s_11(Y), Z)) -> plus_3_out_gaa3(s_11(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
plus_3_out_gaa3(x1, x2, x3)  =  plus_3_out_gaa
if_plus_3_in_1_gaa4(x1, x2, x3, x4)  =  if_plus_3_in_1_gaa1(x4)
IF_PLUS_3_IN_1_GAA4(x1, x2, x3, x4)  =  IF_PLUS_3_IN_1_GAA1(x4)
PLUS_3_IN_GAA3(x1, x2, x3)  =  PLUS_3_IN_GAA1(x1)

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:

PLUS_3_IN_GAA3(s_11(X), Y, Z) -> PLUS_3_IN_GAA3(X, s_11(Y), Z)

The TRS R consists of the following rules:

plus_3_in_gaa3(0_0, Y, Y) -> plus_3_out_gaa3(0_0, Y, Y)
plus_3_in_gaa3(s_11(X), Y, Z) -> if_plus_3_in_1_gaa4(X, Y, Z, plus_3_in_gaa3(X, s_11(Y), Z))
if_plus_3_in_1_gaa4(X, Y, Z, plus_3_out_gaa3(X, s_11(Y), Z)) -> plus_3_out_gaa3(s_11(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
0_0  =  0_0
s_11(x1)  =  s_11(x1)
plus_3_out_gaa3(x1, x2, x3)  =  plus_3_out_gaa
if_plus_3_in_1_gaa4(x1, x2, x3, x4)  =  if_plus_3_in_1_gaa1(x4)
PLUS_3_IN_GAA3(x1, x2, x3)  =  PLUS_3_IN_GAA1(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:

PLUS_3_IN_GAA3(s_11(X), Y, Z) -> PLUS_3_IN_GAA3(X, s_11(Y), Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
PLUS_3_IN_GAA3(x1, x2, x3)  =  PLUS_3_IN_GAA1(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:

PLUS_3_IN_GAA1(s_11(X)) -> PLUS_3_IN_GAA1(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PLUS_3_IN_GAA1}.
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: