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 QDPSizeChangeProof (⇔)
    ↳12 TRUE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇐)
    ↳22 QDP

(0) Obligation:

Clauses:

plus(0, Y, Y).
plus(s(X), Y, Z) :- plus(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)

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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)

(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, Z) → U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)
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

(4) Obligation:

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

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)
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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

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:

PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(Y), Z)

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

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:

PLUS_IN_GAA(s(X)) → PLUS_IN_GAA(X)

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

(11) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • PLUS_IN_GAA(s(X)) → PLUS_IN_GAA(X)
    The graph contains the following edges 1 > 1

(12) TRUE

(13) 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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) 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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)

(15) 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, Z) → U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)
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

(16) Obligation:

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

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_gaa(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)
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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(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, Z) → U1_gaa(X, Y, Z, plus_in_gaa(X, s(Y), Z))
U1_gaa(X, Y, Z, plus_out_gaa(X, s(Y), Z)) → plus_out_gaa(s(X), Y, 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)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) Obligation:

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

PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_GAA(X, s(Y), Z)

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

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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) Obligation:

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

PLUS_IN_GAA(s(X)) → PLUS_IN_GAA(X)

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