Left Termination of the query pattern log_in_2(g, 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 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
          ↳19 MRRProof (⇔)
            ↳20 QDP
            ↳21 DependencyGraphProof (⇔)
            ↳22 TRUE
          ↳23 MRRProof (⇔)
            ↳24 QDP
  ↳25 PrologToPiTRSProof (⇐)
    ↳26 PiTRS
    ↳27 DependencyPairsProof (⇔)
    ↳28 PiDP
    ↳29 DependencyGraphProof (⇔)
    ↳30 AND
      ↳31 PiDP
        ↳32 UsableRulesProof (⇔)
        ↳33 PiDP
        ↳34 PiDPToQDPProof (⇐)
        ↳35 QDP
        ↳36 QDPSizeChangeProof (⇔)
        ↳37 TRUE
      ↳38 PiDP
        ↳39 UsableRulesProof (⇔)
        ↳40 PiDP
        ↳41 PiDPToQDPProof (⇐)
        ↳42 QDP

(0) Obligation:

Clauses:

half(0, 0).
half(s(0), 0).
half(s(s(X)), s(Y)) :- half(X, Y).
log(0, s(0)).
log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y)).

Queries:

log(g,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:
log_in: (b,f)
half_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_ga(x1, x2, x3)  =  U3_ga(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_ga(x1, x2, x3)  =  U3_ga(x3)

(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:

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
LOG_IN_GA(s(X), s(Y)) → HALF_IN_GA(s(X), Z)
HALF_IN_GA(s(s(X)), s(Y)) → U1_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GA(X, Y, half_out_ga(s(X), Z)) → U3_GA(X, Y, log_in_ga(Z, Y))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GA(x1, x2, x3)  =  U3_GA(x3)

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

(4) Obligation:

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

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
LOG_IN_GA(s(X), s(Y)) → HALF_IN_GA(s(X), Z)
HALF_IN_GA(s(s(X)), s(Y)) → U1_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GA(X, Y, half_out_ga(s(X), Z)) → U3_GA(X, Y, log_in_ga(Z, Y))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
U3_GA(x1, x2, x3)  =  U3_GA(x3)

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:

HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_ga(x1, x2, x3)  =  U3_ga(x3)
HALF_IN_GA(x1, x2)  =  HALF_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:

HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)

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

HALF_IN_GA(s(s(X))) → HALF_IN_GA(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:

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

(13) TRUE

(14) Obligation:

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

U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)
LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U3_ga(x1, x2, x3)  =  U3_ga(x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x3)

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_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)
LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))

The TRS R consists of the following rules:

half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0, 0) → half_out_ga(0, 0)

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x3)

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_GA(half_out_ga(Z)) → LOG_IN_GA(Z)
LOG_IN_GA(s(X)) → U2_GA(half_in_ga(s(X)))

The TRS R consists of the following rules:

half_in_ga(s(0)) → half_out_ga(0)
half_in_ga(s(s(X))) → U1_ga(half_in_ga(X))
U1_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)

The set Q consists of the following terms:

half_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 rules of the TRS R:

half_in_ga(s(0)) → half_out_ga(0)
half_in_ga(s(s(X))) → U1_ga(half_in_ga(X))
U1_ga(half_out_ga(Y)) → half_out_ga(s(Y))

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG_IN_GA(x1)) = x1   
POL(U1_ga(x1)) = 3 + x1   
POL(U2_GA(x1)) = x1   
POL(half_in_ga(x1)) = x1   
POL(half_out_ga(x1)) = x1   
POL(s(x1)) = 2 + x1   

(20) Obligation:

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

U2_GA(half_out_ga(Z)) → LOG_IN_GA(Z)
LOG_IN_GA(s(X)) → U2_GA(half_in_ga(s(X)))

The TRS R consists of the following rules:

half_in_ga(0) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U1_ga(x0)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(22) TRUE

(23) 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 rules of the TRS R:

half_in_ga(s(0)) → half_out_ga(0)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 2   
POL(LOG_IN_GA(x1)) = x1   
POL(U1_ga(x1)) = 2·x1   
POL(U2_GA(x1)) = x1   
POL(half_in_ga(x1)) = x1   
POL(half_out_ga(x1)) = x1   
POL(s(x1)) = 2·x1   

(24) Obligation:

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

U2_GA(half_out_ga(Z)) → LOG_IN_GA(Z)
LOG_IN_GA(s(X)) → U2_GA(half_in_ga(s(X)))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U1_ga(half_in_ga(X))
U1_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U1_ga(x0)

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

(25) PrologToPiTRSProof (SOUND transformation)

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

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x1, x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(26) Obligation:

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

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x1, x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)

(27) 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:

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
LOG_IN_GA(s(X), s(Y)) → HALF_IN_GA(s(X), Z)
HALF_IN_GA(s(s(X)), s(Y)) → U1_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GA(X, Y, half_out_ga(s(X), Z)) → U3_GA(X, Y, log_in_ga(Z, Y))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x1, x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U3_GA(x1, x2, x3)  =  U3_GA(x1, x3)

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

(28) Obligation:

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

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
LOG_IN_GA(s(X), s(Y)) → HALF_IN_GA(s(X), Z)
HALF_IN_GA(s(s(X)), s(Y)) → U1_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GA(X, Y, half_out_ga(s(X), Z)) → U3_GA(X, Y, log_in_ga(Z, Y))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x1, x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U3_GA(x1, x2, x3)  =  U3_GA(x1, x3)

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) Complex Obligation (AND)

(31) Obligation:

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

HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x1, x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)

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

(32) UsableRulesProof (EQUIVALENT transformation)

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

(33) Obligation:

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

HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)

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

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

(34) PiDPToQDPProof (SOUND transformation)

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

(35) Obligation:

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

HALF_IN_GA(s(s(X))) → HALF_IN_GA(X)

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

(36) 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:

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

(37) TRUE

(38) Obligation:

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

U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)
LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2)  =  log_in_ga(x1)
0  =  0
log_out_ga(x1, x2)  =  log_out_ga(x1, x2)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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

(39) UsableRulesProof (EQUIVALENT transformation)

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

(40) Obligation:

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

U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)
LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))

The TRS R consists of the following rules:

half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0, 0) → half_out_ga(0, 0)

The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
LOG_IN_GA(x1, x2)  =  LOG_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)

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

(41) PiDPToQDPProof (SOUND transformation)

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

(42) Obligation:

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

U2_GA(X, half_out_ga(s(X), Z)) → LOG_IN_GA(Z)
LOG_IN_GA(s(X)) → U2_GA(X, half_in_ga(s(X)))

The TRS R consists of the following rules:

half_in_ga(s(0)) → half_out_ga(s(0), 0)
half_in_ga(s(s(X))) → U1_ga(X, half_in_ga(X))
U1_ga(X, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0) → half_out_ga(0, 0)

The set Q consists of the following terms:

half_in_ga(x0)
U1_ga(x0, x1)

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