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

0 Prolog
↳1 PrologToPiTRSProof (⇒, 0 ms)
↳2 PiTRS
↳3 DependencyPairsProof (⇔, 16 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇒, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 PiDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇒, 0 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 28 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

Clauses:

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

Query: plus(g,a,a)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus_in: (b,f,f)
p_in: (b,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)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, 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)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_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, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, 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)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_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, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_ga(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_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)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, 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)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_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, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_ga(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_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)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, 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)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(s(X), s(Y))

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)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, 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)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(s(X), s(Y))

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

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

P_IN_GA(s(s(X))) → P_IN_GA(s(X))

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

(12) 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:

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

(13) YES

(14) Obligation:

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_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)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, 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)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)

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

(17) PiDPToQDPProof (SOUND transformation)

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

(18) Obligation:

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

U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s(X)) → U2_GAA(p_in_ga(s(X)))

The TRS R consists of the following rules:

p_in_ga(s(0)) → p_out_ga(0)
p_in_ga(s(s(X))) → U1_ga(p_in_ga(s(X)))
U1_ga(p_out_ga(s(Y))) → p_out_ga(s(s(Y)))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)

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

(19) 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:

U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s(X)) → U2_GAA(p_in_ga(s(X)))

Strictly oriented rules of the TRS R:

p_in_ga(s(0)) → p_out_ga(0)
p_in_ga(s(s(X))) → U1_ga(p_in_ga(s(X)))
U1_ga(p_out_ga(s(Y))) → p_out_ga(s(s(Y)))

Used ordering: Knuth-Bendix order [KBO] with precedence:
0 > s1 > pinga1 > U1ga1 > U2GAA1 > PLUSINGAA1 > poutga1

and weight map:

0=1
p_in_ga_1=1
s_1=4
p_out_ga_1=3
U1_ga_1=4
U2_GAA_1=1
PLUS_IN_GAA_1=3

The variable weight is 1

(20) Obligation:

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

p_in_ga(x0)
U1_ga(x0)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) YES