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



↳ PROLOG
  ↳ PrologToPiTRSProof

p2(s1(X), X).
plus3(00, Y, Y).
plus3(s1(X), Y, s1(Z)) :- p2(s1(X), U), plus3(U, 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, s_11(Z)) -> if_plus_3_in_1_gaa4(X, Y, Z, p_2_in_ga2(s_11(X), U))
p_2_in_ga2(s_11(X), X) -> p_2_out_ga2(s_11(X), X)
if_plus_3_in_1_gaa4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_out_gaa3(U, Y, Z)) -> plus_3_out_gaa3(s_11(X), Y, s_11(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
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)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_plus_3_in_2_gaa5(x1, x2, x3, x4, x5)  =  if_plus_3_in_2_gaa1(x5)

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, s_11(Z)) -> if_plus_3_in_1_gaa4(X, Y, Z, p_2_in_ga2(s_11(X), U))
p_2_in_ga2(s_11(X), X) -> p_2_out_ga2(s_11(X), X)
if_plus_3_in_1_gaa4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_out_gaa3(U, Y, Z)) -> plus_3_out_gaa3(s_11(X), Y, s_11(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
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)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_plus_3_in_2_gaa5(x1, x2, x3, x4, x5)  =  if_plus_3_in_2_gaa1(x5)


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

PLUS_3_IN_GAA3(s_11(X), Y, s_11(Z)) -> IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_in_ga2(s_11(X), U))
PLUS_3_IN_GAA3(s_11(X), Y, s_11(Z)) -> P_2_IN_GA2(s_11(X), U)
IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> IF_PLUS_3_IN_2_GAA5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> PLUS_3_IN_GAA3(U, 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, s_11(Z)) -> if_plus_3_in_1_gaa4(X, Y, Z, p_2_in_ga2(s_11(X), U))
p_2_in_ga2(s_11(X), X) -> p_2_out_ga2(s_11(X), X)
if_plus_3_in_1_gaa4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_out_gaa3(U, Y, Z)) -> plus_3_out_gaa3(s_11(X), Y, s_11(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
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)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_plus_3_in_2_gaa5(x1, x2, x3, x4, x5)  =  if_plus_3_in_2_gaa1(x5)
IF_PLUS_3_IN_1_GAA4(x1, x2, x3, x4)  =  IF_PLUS_3_IN_1_GAA1(x4)
P_2_IN_GA2(x1, x2)  =  P_2_IN_GA1(x1)
PLUS_3_IN_GAA3(x1, x2, x3)  =  PLUS_3_IN_GAA1(x1)
IF_PLUS_3_IN_2_GAA5(x1, x2, x3, x4, x5)  =  IF_PLUS_3_IN_2_GAA1(x5)

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, s_11(Z)) -> IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_in_ga2(s_11(X), U))
PLUS_3_IN_GAA3(s_11(X), Y, s_11(Z)) -> P_2_IN_GA2(s_11(X), U)
IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> IF_PLUS_3_IN_2_GAA5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> PLUS_3_IN_GAA3(U, 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, s_11(Z)) -> if_plus_3_in_1_gaa4(X, Y, Z, p_2_in_ga2(s_11(X), U))
p_2_in_ga2(s_11(X), X) -> p_2_out_ga2(s_11(X), X)
if_plus_3_in_1_gaa4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_out_gaa3(U, Y, Z)) -> plus_3_out_gaa3(s_11(X), Y, s_11(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
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)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_plus_3_in_2_gaa5(x1, x2, x3, x4, x5)  =  if_plus_3_in_2_gaa1(x5)
IF_PLUS_3_IN_1_GAA4(x1, x2, x3, x4)  =  IF_PLUS_3_IN_1_GAA1(x4)
P_2_IN_GA2(x1, x2)  =  P_2_IN_GA1(x1)
PLUS_3_IN_GAA3(x1, x2, x3)  =  PLUS_3_IN_GAA1(x1)
IF_PLUS_3_IN_2_GAA5(x1, x2, x3, x4, x5)  =  IF_PLUS_3_IN_2_GAA1(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> PLUS_3_IN_GAA3(U, Y, Z)
PLUS_3_IN_GAA3(s_11(X), Y, s_11(Z)) -> IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_in_ga2(s_11(X), U))

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, s_11(Z)) -> if_plus_3_in_1_gaa4(X, Y, Z, p_2_in_ga2(s_11(X), U))
p_2_in_ga2(s_11(X), X) -> p_2_out_ga2(s_11(X), X)
if_plus_3_in_1_gaa4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_in_gaa3(U, Y, Z))
if_plus_3_in_2_gaa5(X, Y, Z, U, plus_3_out_gaa3(U, Y, Z)) -> plus_3_out_gaa3(s_11(X), Y, s_11(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa3(x1, x2, x3)  =  plus_3_in_gaa1(x1)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
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)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
if_plus_3_in_2_gaa5(x1, x2, x3, x4, x5)  =  if_plus_3_in_2_gaa1(x5)
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:

IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_out_ga2(s_11(X), U)) -> PLUS_3_IN_GAA3(U, Y, Z)
PLUS_3_IN_GAA3(s_11(X), Y, s_11(Z)) -> IF_PLUS_3_IN_1_GAA4(X, Y, Z, p_2_in_ga2(s_11(X), U))

The TRS R consists of the following rules:

p_2_in_ga2(s_11(X), X) -> p_2_out_ga2(s_11(X), X)

The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
p_2_in_ga2(x1, x2)  =  p_2_in_ga1(x1)
p_2_out_ga2(x1, x2)  =  p_2_out_ga1(x2)
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
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
                      ↳ RuleRemovalProof

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

IF_PLUS_3_IN_1_GAA1(p_2_out_ga1(U)) -> PLUS_3_IN_GAA1(U)
PLUS_3_IN_GAA1(s_11(X)) -> IF_PLUS_3_IN_1_GAA1(p_2_in_ga1(s_11(X)))

The TRS R consists of the following rules:

p_2_in_ga1(s_11(X)) -> p_2_out_ga1(X)

The set Q consists of the following terms:

p_2_in_ga1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PLUS_3_IN_GAA1, IF_PLUS_3_IN_1_GAA1}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF_PLUS_3_IN_1_GAA1(p_2_out_ga1(U)) -> PLUS_3_IN_GAA1(U)

Strictly oriented rules of the TRS R:

p_2_in_ga1(s_11(X)) -> p_2_out_ga1(X)

Used ordering: POLO with Polynomial interpretation:

POL(p_2_out_ga1(x1)) = 1 + x1   
POL(p_2_in_ga1(x1)) = x1   
POL(s_11(x1)) = 2 + x1   
POL(IF_PLUS_3_IN_1_GAA1(x1)) = x1   
POL(PLUS_3_IN_GAA1(x1)) = x1   



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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

PLUS_3_IN_GAA1(s_11(X)) -> IF_PLUS_3_IN_1_GAA1(p_2_in_ga1(s_11(X)))

R is empty.
The set Q consists of the following terms:

p_2_in_ga1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PLUS_3_IN_1_GAA1, PLUS_3_IN_GAA1}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.