Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
    ↳11 MRRProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 PrologToPiTRSProof (⇐)
    ↳16 PiTRS
    ↳17 DependencyPairsProof (⇔)
    ↳18 PiDP
    ↳19 DependencyGraphProof (⇔)
    ↳20 PiDP
    ↳21 UsableRulesProof (⇔)
    ↳22 PiDP
    ↳23 PiDPToQDPProof (⇐)
    ↳24 QDP

(0) Obligation:

Clauses:

p(s(X), X).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).

Queries:

plus(g,a,a).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → U2_GAA(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x5)

We have to consider all (P,R,Pi)-chains

(4) Obligation:

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

PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → U2_GAA(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x5)

We have to consider all (P,R,Pi)-chains

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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

U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)

We have to consider all (P,R,Pi)-chains

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

p_in_ga(s(X), X) → p_out_ga(s(X), X)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)

We have to consider all (P,R,Pi)-chains

(9) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

U1_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s(X)) → U1_GAA(p_in_ga(s(X)))

The TRS R consists of the following rules:

p_in_ga(s(X)) → p_out_ga(X)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.

(11) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U1_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s(X)) → U1_GAA(p_in_ga(s(X)))

Strictly oriented rules of the TRS R:

p_in_ga(s(X)) → p_out_ga(X)

Used ordering: Polynomial interpretation [POLO]:

POL(PLUS_IN_GAA(x1)) = 1 + x1   
POL(U1_GAA(x1)) = x1   
POL(p_in_ga(x1)) = x1   
POL(p_out_ga(x1)) = 2 + x1   
POL(s(x1)) = 3 + x1   

(12) Obligation:

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

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x5)

(17) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → U2_GAA(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x5)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x1, x5)

We have to consider all (P,R,Pi)-chains

(18) Obligation:

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

PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → U2_GAA(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x5)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x1, x5)

We have to consider all (P,R,Pi)-chains

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(20) Obligation:

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

U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U1_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U1_gaa(X, Y, Z, p_out_ga(s(X), U)) → U2_gaa(X, Y, Z, U, plus_in_gaa(U, Y, Z))
U2_gaa(X, Y, Z, U, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s(x1)
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x1, x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x5)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)

We have to consider all (P,R,Pi)-chains

(21) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(22) Obligation:

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

U1_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U1_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

p_in_ga(s(X), X) → p_out_ga(s(X), X)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x1, x4)

We have to consider all (P,R,Pi)-chains

(23) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(24) Obligation:

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

U1_GAA(X, p_out_ga(s(X), U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s(X)) → U1_GAA(X, p_in_ga(s(X)))

The TRS R consists of the following rules:

p_in_ga(s(X)) → p_out_ga(s(X), X)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.