YES Left Termination proof of ../tpdb/LP/lpexamples/log2a.pl
Left Termination of the query pattern log2_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

log2(X, Y) :- log2(X, 0, Y).
log2(0, I, I).
log2(s(0), I, I).
log2(s(s(X)), I, Y) :- ','(half(s(s(X)), X1), log2(X1, s(I), Y)).
half(0, 0).
half(s(0), 0).
half(s(s(X)), s(Y)) :- half(X, Y).

Queries:

log2(g,a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
log2_in: (b,f)
log2_in: (b,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:

log2_in_ga(X, Y) → U1_ga(X, Y, log2_in_gga(X, 0, Y))
log2_in_gga(0, I, I) → log2_out_gga(0, I, I)
log2_in_gga(s(0), I, I) → log2_out_gga(s(0), I, I)
log2_in_gga(s(s(X)), I, Y) → U2_gga(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y))
U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) → log2_out_gga(s(s(X)), I, Y)
U1_ga(X, Y, log2_out_gga(X, 0, Y)) → log2_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ga(x1, x2)  =  log2_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
log2_in_gga(x1, x2, x3)  =  log2_in_gga(x1, x2)
0  =  0
log2_out_gga(x1, x2, x3)  =  log2_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x2, x4)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
log2_out_ga(x1, x2)  =  log2_out_ga(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

log2_in_ga(X, Y) → U1_ga(X, Y, log2_in_gga(X, 0, Y))
log2_in_gga(0, I, I) → log2_out_gga(0, I, I)
log2_in_gga(s(0), I, I) → log2_out_gga(s(0), I, I)
log2_in_gga(s(s(X)), I, Y) → U2_gga(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y))
U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) → log2_out_gga(s(s(X)), I, Y)
U1_ga(X, Y, log2_out_gga(X, 0, Y)) → log2_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ga(x1, x2)  =  log2_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
log2_in_gga(x1, x2, x3)  =  log2_in_gga(x1, x2)
0  =  0
log2_out_gga(x1, x2, x3)  =  log2_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x2, x4)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
log2_out_ga(x1, x2)  =  log2_out_ga(x2)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

LOG2_IN_GA(X, Y) → U1_GA(X, Y, log2_in_gga(X, 0, Y))
LOG2_IN_GA(X, Y) → LOG2_IN_GGA(X, 0, Y)
LOG2_IN_GGA(s(s(X)), I, Y) → U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1))
LOG2_IN_GGA(s(s(X)), I, Y) → HALF_IN_GA(s(s(X)), X1)
HALF_IN_GA(s(s(X)), s(Y)) → U4_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_GGA(X, I, Y, log2_in_gga(X1, s(I), Y))
U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGA(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ga(X, Y) → U1_ga(X, Y, log2_in_gga(X, 0, Y))
log2_in_gga(0, I, I) → log2_out_gga(0, I, I)
log2_in_gga(s(0), I, I) → log2_out_gga(s(0), I, I)
log2_in_gga(s(s(X)), I, Y) → U2_gga(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y))
U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) → log2_out_gga(s(s(X)), I, Y)
U1_ga(X, Y, log2_out_gga(X, 0, Y)) → log2_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ga(x1, x2)  =  log2_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
log2_in_gga(x1, x2, x3)  =  log2_in_gga(x1, x2)
0  =  0
log2_out_gga(x1, x2, x3)  =  log2_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x2, x4)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
log2_out_ga(x1, x2)  =  log2_out_ga(x2)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)
U4_GA(x1, x2, x3)  =  U4_GA(x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x2, x4)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
LOG2_IN_GGA(x1, x2, x3)  =  LOG2_IN_GGA(x1, x2)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x4)
LOG2_IN_GA(x1, x2)  =  LOG2_IN_GA(x1)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

LOG2_IN_GA(X, Y) → U1_GA(X, Y, log2_in_gga(X, 0, Y))
LOG2_IN_GA(X, Y) → LOG2_IN_GGA(X, 0, Y)
LOG2_IN_GGA(s(s(X)), I, Y) → U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1))
LOG2_IN_GGA(s(s(X)), I, Y) → HALF_IN_GA(s(s(X)), X1)
HALF_IN_GA(s(s(X)), s(Y)) → U4_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_GGA(X, I, Y, log2_in_gga(X1, s(I), Y))
U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGA(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ga(X, Y) → U1_ga(X, Y, log2_in_gga(X, 0, Y))
log2_in_gga(0, I, I) → log2_out_gga(0, I, I)
log2_in_gga(s(0), I, I) → log2_out_gga(s(0), I, I)
log2_in_gga(s(s(X)), I, Y) → U2_gga(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y))
U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) → log2_out_gga(s(s(X)), I, Y)
U1_ga(X, Y, log2_out_gga(X, 0, Y)) → log2_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ga(x1, x2)  =  log2_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
log2_in_gga(x1, x2, x3)  =  log2_in_gga(x1, x2)
0  =  0
log2_out_gga(x1, x2, x3)  =  log2_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x2, x4)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
log2_out_ga(x1, x2)  =  log2_out_ga(x2)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)
U4_GA(x1, x2, x3)  =  U4_GA(x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x2, x4)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
LOG2_IN_GGA(x1, x2, x3)  =  LOG2_IN_GGA(x1, x2)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x4)
LOG2_IN_GA(x1, x2)  =  LOG2_IN_GA(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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:

log2_in_ga(X, Y) → U1_ga(X, Y, log2_in_gga(X, 0, Y))
log2_in_gga(0, I, I) → log2_out_gga(0, I, I)
log2_in_gga(s(0), I, I) → log2_out_gga(s(0), I, I)
log2_in_gga(s(s(X)), I, Y) → U2_gga(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y))
U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) → log2_out_gga(s(s(X)), I, Y)
U1_ga(X, Y, log2_out_gga(X, 0, Y)) → log2_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ga(x1, x2)  =  log2_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
log2_in_gga(x1, x2, x3)  =  log2_in_gga(x1, x2)
0  =  0
log2_out_gga(x1, x2, x3)  =  log2_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x2, x4)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
log2_out_ga(x1, x2)  =  log2_out_ga(x2)
HALF_IN_GA(x1, x2)  =  HALF_IN_GA(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof

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

LOG2_IN_GGA(s(s(X)), I, Y) → U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1))
U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGA(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ga(X, Y) → U1_ga(X, Y, log2_in_gga(X, 0, Y))
log2_in_gga(0, I, I) → log2_out_gga(0, I, I)
log2_in_gga(s(0), I, I) → log2_out_gga(s(0), I, I)
log2_in_gga(s(s(X)), I, Y) → U2_gga(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_gga(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_gga(X, I, Y, log2_in_gga(X1, s(I), Y))
U3_gga(X, I, Y, log2_out_gga(X1, s(I), Y)) → log2_out_gga(s(s(X)), I, Y)
U1_ga(X, Y, log2_out_gga(X, 0, Y)) → log2_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ga(x1, x2)  =  log2_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
log2_in_gga(x1, x2, x3)  =  log2_in_gga(x1, x2)
0  =  0
log2_out_gga(x1, x2, x3)  =  log2_out_gga(x3)
s(x1)  =  s(x1)
U2_gga(x1, x2, x3, x4)  =  U2_gga(x2, x4)
half_in_ga(x1, x2)  =  half_in_ga(x1)
half_out_ga(x1, x2)  =  half_out_ga(x2)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U3_gga(x1, x2, x3, x4)  =  U3_gga(x4)
log2_out_ga(x1, x2)  =  log2_out_ga(x2)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x2, x4)
LOG2_IN_GGA(x1, x2, x3)  =  LOG2_IN_GGA(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

LOG2_IN_GGA(s(s(X)), I, Y) → U2_GGA(X, I, Y, half_in_ga(s(s(X)), X1))
U2_GGA(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGA(X1, s(I), Y)

The TRS R consists of the following rules:

half_in_ga(s(s(X)), s(Y)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_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)
half_in_ga(s(0), 0) → half_out_ga(s(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)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x2, x4)
LOG2_IN_GGA(x1, x2, x3)  =  LOG2_IN_GGA(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

LOG2_IN_GGA(s(s(X)), I) → U2_GGA(I, half_in_ga(s(s(X))))
U2_GGA(I, half_out_ga(X1)) → LOG2_IN_GGA(X1, s(I))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(half_in_ga(X))
U4_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)
half_in_ga(s(0)) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U4_ga(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U2_GGA(I, half_out_ga(X1)) → LOG2_IN_GGA(X1, s(I))

Strictly oriented rules of the TRS R:

half_in_ga(s(0)) → half_out_ga(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(LOG2_IN_GGA(x1, x2)) = 2·x1 + x2   
POL(U2_GGA(x1, x2)) = 2 + x1 + x2   
POL(U4_ga(x1)) = 2 + x1   
POL(half_in_ga(x1)) = x1   
POL(half_out_ga(x1)) = 2·x1   
POL(s(x1)) = 1 + x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ DependencyGraphProof

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

LOG2_IN_GGA(s(s(X)), I) → U2_GGA(I, half_in_ga(s(s(X))))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(half_in_ga(X))
U4_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)
U4_ga(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.