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

Queries:

plus(g,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

p23(s(T29), s(X53)) :- p23(T29, X53).
plus1(0, T5, T5).
plus1(s(0), T20, s(T20)).
plus1(s(s(T23)), T12, s(T13)) :- p23(T23, X36).
plus1(s(s(T23)), T12, s(T13)) :- ','(p23(T23, T25), plus1(s(s(T25)), T12, T13)).

Queries:

plus1(g,a,a).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
plus1_in: (b,f,f)
p23_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T20, s(T20)) → plus1_out_gaa(s(0), T20, s(T20))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, p23_in_ga(T23, X36))
p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))
U2_gaa(T23, T12, T13, p23_out_ga(T23, X36)) → plus1_out_gaa(s(s(T23)), T12, s(T13))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, p23_in_ga(T23, T25))
U3_gaa(T23, T12, T13, p23_out_ga(T23, T25)) → U4_gaa(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U4_gaa(T23, T12, T13, plus1_out_gaa(s(s(T25)), T12, T13)) → plus1_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T20, s(T20)) → plus1_out_gaa(s(0), T20, s(T20))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, p23_in_ga(T23, X36))
p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))
U2_gaa(T23, T12, T13, p23_out_ga(T23, X36)) → plus1_out_gaa(s(s(T23)), T12, s(T13))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, p23_in_ga(T23, T25))
U3_gaa(T23, T12, T13, p23_out_ga(T23, T25)) → U4_gaa(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U4_gaa(T23, T12, T13, plus1_out_gaa(s(s(T25)), T12, T13)) → plus1_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x4)

(5) 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:

PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → U2_GAA(T23, T12, T13, p23_in_ga(T23, X36))
PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → P23_IN_GA(T23, X36)
P23_IN_GA(s(T29), s(X53)) → U1_GA(T29, X53, p23_in_ga(T29, X53))
P23_IN_GA(s(T29), s(X53)) → P23_IN_GA(T29, X53)
PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, p23_in_ga(T23, T25))
U3_GAA(T23, T12, T13, p23_out_ga(T23, T25)) → U4_GAA(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U3_GAA(T23, T12, T13, p23_out_ga(T23, T25)) → PLUS1_IN_GAA(s(s(T25)), T12, T13)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T20, s(T20)) → plus1_out_gaa(s(0), T20, s(T20))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, p23_in_ga(T23, X36))
p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))
U2_gaa(T23, T12, T13, p23_out_ga(T23, X36)) → plus1_out_gaa(s(s(T23)), T12, s(T13))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, p23_in_ga(T23, T25))
U3_gaa(T23, T12, T13, p23_out_ga(T23, T25)) → U4_gaa(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U4_gaa(T23, T12, T13, plus1_out_gaa(s(s(T25)), T12, T13)) → plus1_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x4)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
P23_IN_GA(x1, x2)  =  P23_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x4)

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

(6) Obligation:

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

PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → U2_GAA(T23, T12, T13, p23_in_ga(T23, X36))
PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → P23_IN_GA(T23, X36)
P23_IN_GA(s(T29), s(X53)) → U1_GA(T29, X53, p23_in_ga(T29, X53))
P23_IN_GA(s(T29), s(X53)) → P23_IN_GA(T29, X53)
PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, p23_in_ga(T23, T25))
U3_GAA(T23, T12, T13, p23_out_ga(T23, T25)) → U4_GAA(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U3_GAA(T23, T12, T13, p23_out_ga(T23, T25)) → PLUS1_IN_GAA(s(s(T25)), T12, T13)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T20, s(T20)) → plus1_out_gaa(s(0), T20, s(T20))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, p23_in_ga(T23, X36))
p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))
U2_gaa(T23, T12, T13, p23_out_ga(T23, X36)) → plus1_out_gaa(s(s(T23)), T12, s(T13))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, p23_in_ga(T23, T25))
U3_gaa(T23, T12, T13, p23_out_ga(T23, T25)) → U4_gaa(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U4_gaa(T23, T12, T13, plus1_out_gaa(s(s(T25)), T12, T13)) → plus1_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x4)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4)  =  U2_GAA(x4)
P23_IN_GA(x1, x2)  =  P23_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)
U4_GAA(x1, x2, x3, x4)  =  U4_GAA(x4)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

P23_IN_GA(s(T29), s(X53)) → P23_IN_GA(T29, X53)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T20, s(T20)) → plus1_out_gaa(s(0), T20, s(T20))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, p23_in_ga(T23, X36))
p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))
U2_gaa(T23, T12, T13, p23_out_ga(T23, X36)) → plus1_out_gaa(s(s(T23)), T12, s(T13))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, p23_in_ga(T23, T25))
U3_gaa(T23, T12, T13, p23_out_ga(T23, T25)) → U4_gaa(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U4_gaa(T23, T12, T13, plus1_out_gaa(s(s(T25)), T12, T13)) → plus1_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x4)
P23_IN_GA(x1, x2)  =  P23_IN_GA(x1)

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

P23_IN_GA(s(T29), s(X53)) → P23_IN_GA(T29, X53)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
P23_IN_GA(x1, x2)  =  P23_IN_GA(x1)

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

(12) PiDPToQDPProof (SOUND transformation)

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

(13) Obligation:

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

P23_IN_GA(s(T29)) → P23_IN_GA(T29)

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

(14) 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:

  • P23_IN_GA(s(T29)) → P23_IN_GA(T29)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, p23_in_ga(T23, T25))
U3_GAA(T23, T12, T13, p23_out_ga(T23, T25)) → PLUS1_IN_GAA(s(s(T25)), T12, T13)

The TRS R consists of the following rules:

plus1_in_gaa(0, T5, T5) → plus1_out_gaa(0, T5, T5)
plus1_in_gaa(s(0), T20, s(T20)) → plus1_out_gaa(s(0), T20, s(T20))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, p23_in_ga(T23, X36))
p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))
U2_gaa(T23, T12, T13, p23_out_ga(T23, X36)) → plus1_out_gaa(s(s(T23)), T12, s(T13))
plus1_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, p23_in_ga(T23, T25))
U3_gaa(T23, T12, T13, p23_out_ga(T23, T25)) → U4_gaa(T23, T12, T13, plus1_in_gaa(s(s(T25)), T12, T13))
U4_gaa(T23, T12, T13, plus1_out_gaa(s(s(T25)), T12, T13)) → plus1_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plus1_in_gaa(x1, x2, x3)  =  plus1_in_gaa(x1)
0  =  0
plus1_out_gaa(x1, x2, x3)  =  plus1_out_gaa
s(x1)  =  s(x1)
U2_gaa(x1, x2, x3, x4)  =  U2_gaa(x4)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
U3_gaa(x1, x2, x3, x4)  =  U3_gaa(x4)
U4_gaa(x1, x2, x3, x4)  =  U4_gaa(x4)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)

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

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

PLUS1_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, p23_in_ga(T23, T25))
U3_GAA(T23, T12, T13, p23_out_ga(T23, T25)) → PLUS1_IN_GAA(s(s(T25)), T12, T13)

The TRS R consists of the following rules:

p23_in_ga(s(T29), s(X53)) → U1_ga(T29, X53, p23_in_ga(T29, X53))
U1_ga(T29, X53, p23_out_ga(T29, X53)) → p23_out_ga(s(T29), s(X53))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
p23_in_ga(x1, x2)  =  p23_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
p23_out_ga(x1, x2)  =  p23_out_ga(x2)
PLUS1_IN_GAA(x1, x2, x3)  =  PLUS1_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4)  =  U3_GAA(x4)

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

(19) PiDPToQDPProof (SOUND transformation)

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

(20) Obligation:

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

PLUS1_IN_GAA(s(s(T23))) → U3_GAA(p23_in_ga(T23))
U3_GAA(p23_out_ga(T25)) → PLUS1_IN_GAA(s(s(T25)))

The TRS R consists of the following rules:

p23_in_ga(s(T29)) → U1_ga(p23_in_ga(T29))
U1_ga(p23_out_ga(X53)) → p23_out_ga(s(X53))

The set Q consists of the following terms:

p23_in_ga(x0)
U1_ga(x0)

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

(21) 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:

PLUS1_IN_GAA(s(s(T23))) → U3_GAA(p23_in_ga(T23))
U3_GAA(p23_out_ga(T25)) → PLUS1_IN_GAA(s(s(T25)))


Used ordering: Polynomial interpretation [POLO]:

POL(PLUS1_IN_GAA(x1)) = 1 + x1   
POL(U1_ga(x1)) = x1   
POL(U3_GAA(x1)) = x1   
POL(p23_in_ga(x1)) = x1   
POL(p23_out_ga(x1)) = 2 + x1   
POL(s(x1)) = x1   

(22) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

p23_in_ga(s(T29)) → U1_ga(p23_in_ga(T29))
U1_ga(p23_out_ga(X53)) → p23_out_ga(s(X53))

The set Q consists of the following terms:

p23_in_ga(x0)
U1_ga(x0)

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) YES