Left Termination of the query pattern fl_in_3(a, g, a) w.r.t. the given Prolog program could not be shown:

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
        ↳19 NonTerminationProof (⇔)
        ↳20 NO
    ↳21 PiDP
        ↳22 PiDPToQDPProof (⇐)
        ↳23 QDP
        ↳24 QDPOrderProof (⇔)
        ↳25 QDP
        ↳26 DependencyGraphProof (⇔)
        ↳27 QDP
        ↳28 UsableRulesProof (⇔)
        ↳29 QDP
        ↳30 QReductionProof (⇔)
        ↳31 QDP
        ↳32 NonTerminationProof (⇔)
        ↳33 NO

(0) Obligation:

Clauses:

fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).

Queries:

fl(a,g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

p6([], [], .(T24, T25), s(T26)) :- p6(T24, X38, T25, T26).
append31(.(T77, T80), X128, .(T77, T79)) :- append31(T80, X128, T79).
fl1(.(T10, T11), [], s(T12)) :- p6(T10, X9, T11, T12).
fl1(.([], T45), T44, s(T46)) :- fl1(T45, T44, T46).
fl1(.(.(T53, T56), T57), .(T53, T55), s(T58)) :- append31(T56, X90, T55).
fl1(.(.(T53, T56), T62), .(T53, T55), s(T63)) :- ','(appendc31(T56, T61, T55), fl1(T62, T61, T63)).

Clauses:

qc6([], [], [], 0).
qc6([], [], .(T24, T25), s(T26)) :- qc6(T24, X38, T25, T26).
flc1([], [], 0).
flc1(.(T10, T11), [], s(T12)) :- qc6(T10, X9, T11, T12).
flc1(.([], T45), T44, s(T46)) :- flc1(T45, T44, T46).
flc1(.(.(T53, T56), T62), .(T53, T55), s(T63)) :- ','(appendc31(T56, T61, T55), flc1(T62, T61, T63)).
appendc31([], T70, T70).
appendc31(.(T77, T80), X128, .(T77, T79)) :- appendc31(T80, X128, T79).

Afs:

fl1(x1, x2, x3)  =  fl1(x2)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
fl1_in: (f,b,f)
p6_in: (f,f,f,f)
append31_in: (f,f,b)
appendc31_in: (f,f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

FL1_IN_AGA(.(T10, T11), [], s(T12)) → U3_AGA(T10, T11, T12, p6_in_aaaa(T10, X9, T11, T12))
FL1_IN_AGA(.(T10, T11), [], s(T12)) → P6_IN_AAAA(T10, X9, T11, T12)
P6_IN_AAAA([], [], .(T24, T25), s(T26)) → U1_AAAA(T24, T25, T26, p6_in_aaaa(T24, X38, T25, T26))
P6_IN_AAAA([], [], .(T24, T25), s(T26)) → P6_IN_AAAA(T24, X38, T25, T26)
FL1_IN_AGA(.([], T45), T44, s(T46)) → U4_AGA(T45, T44, T46, fl1_in_aga(T45, T44, T46))
FL1_IN_AGA(.([], T45), T44, s(T46)) → FL1_IN_AGA(T45, T44, T46)
FL1_IN_AGA(.(.(T53, T56), T57), .(T53, T55), s(T58)) → U5_AGA(T53, T56, T57, T55, T58, append31_in_aag(T56, X90, T55))
FL1_IN_AGA(.(.(T53, T56), T57), .(T53, T55), s(T58)) → APPEND31_IN_AAG(T56, X90, T55)
APPEND31_IN_AAG(.(T77, T80), X128, .(T77, T79)) → U2_AAG(T77, T80, X128, T79, append31_in_aag(T80, X128, T79))
APPEND31_IN_AAG(.(T77, T80), X128, .(T77, T79)) → APPEND31_IN_AAG(T80, X128, T79)
FL1_IN_AGA(.(.(T53, T56), T62), .(T53, T55), s(T63)) → U6_AGA(T53, T56, T62, T55, T63, appendc31_in_aag(T56, T61, T55))
U6_AGA(T53, T56, T62, T55, T63, appendc31_out_aag(T56, T61, T55)) → U7_AGA(T53, T56, T62, T55, T63, fl1_in_aga(T62, T61, T63))
U6_AGA(T53, T56, T62, T55, T63, appendc31_out_aag(T56, T61, T55)) → FL1_IN_AGA(T62, T61, T63)

The TRS R consists of the following rules:

appendc31_in_aag([], T70, T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T80), X128, .(T77, T79)) → U14_aag(T77, T80, X128, T79, appendc31_in_aag(T80, X128, T79))
U14_aag(T77, T80, X128, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The argument filtering Pi contains the following mapping:
fl1_in_aga(x1, x2, x3)  =  fl1_in_aga(x2)
[]  =  []
p6_in_aaaa(x1, x2, x3, x4)  =  p6_in_aaaa
.(x1, x2)  =  .(x1, x2)
append31_in_aag(x1, x2, x3)  =  append31_in_aag(x3)
appendc31_in_aag(x1, x2, x3)  =  appendc31_in_aag(x3)
appendc31_out_aag(x1, x2, x3)  =  appendc31_out_aag(x1, x2, x3)
U14_aag(x1, x2, x3, x4, x5)  =  U14_aag(x1, x4, x5)
s(x1)  =  s(x1)
FL1_IN_AGA(x1, x2, x3)  =  FL1_IN_AGA(x2)
U3_AGA(x1, x2, x3, x4)  =  U3_AGA(x4)
P6_IN_AAAA(x1, x2, x3, x4)  =  P6_IN_AAAA
U1_AAAA(x1, x2, x3, x4)  =  U1_AAAA(x4)
U4_AGA(x1, x2, x3, x4)  =  U4_AGA(x2, x4)
U5_AGA(x1, x2, x3, x4, x5, x6)  =  U5_AGA(x1, x4, x6)
APPEND31_IN_AAG(x1, x2, x3)  =  APPEND31_IN_AAG(x3)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x4, x5)
U6_AGA(x1, x2, x3, x4, x5, x6)  =  U6_AGA(x1, x4, x6)
U7_AGA(x1, x2, x3, x4, x5, x6)  =  U7_AGA(x1, x4, x6)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

FL1_IN_AGA(.(T10, T11), [], s(T12)) → U3_AGA(T10, T11, T12, p6_in_aaaa(T10, X9, T11, T12))
FL1_IN_AGA(.(T10, T11), [], s(T12)) → P6_IN_AAAA(T10, X9, T11, T12)
P6_IN_AAAA([], [], .(T24, T25), s(T26)) → U1_AAAA(T24, T25, T26, p6_in_aaaa(T24, X38, T25, T26))
P6_IN_AAAA([], [], .(T24, T25), s(T26)) → P6_IN_AAAA(T24, X38, T25, T26)
FL1_IN_AGA(.([], T45), T44, s(T46)) → U4_AGA(T45, T44, T46, fl1_in_aga(T45, T44, T46))
FL1_IN_AGA(.([], T45), T44, s(T46)) → FL1_IN_AGA(T45, T44, T46)
FL1_IN_AGA(.(.(T53, T56), T57), .(T53, T55), s(T58)) → U5_AGA(T53, T56, T57, T55, T58, append31_in_aag(T56, X90, T55))
FL1_IN_AGA(.(.(T53, T56), T57), .(T53, T55), s(T58)) → APPEND31_IN_AAG(T56, X90, T55)
APPEND31_IN_AAG(.(T77, T80), X128, .(T77, T79)) → U2_AAG(T77, T80, X128, T79, append31_in_aag(T80, X128, T79))
APPEND31_IN_AAG(.(T77, T80), X128, .(T77, T79)) → APPEND31_IN_AAG(T80, X128, T79)
FL1_IN_AGA(.(.(T53, T56), T62), .(T53, T55), s(T63)) → U6_AGA(T53, T56, T62, T55, T63, appendc31_in_aag(T56, T61, T55))
U6_AGA(T53, T56, T62, T55, T63, appendc31_out_aag(T56, T61, T55)) → U7_AGA(T53, T56, T62, T55, T63, fl1_in_aga(T62, T61, T63))
U6_AGA(T53, T56, T62, T55, T63, appendc31_out_aag(T56, T61, T55)) → FL1_IN_AGA(T62, T61, T63)

The TRS R consists of the following rules:

appendc31_in_aag([], T70, T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T80), X128, .(T77, T79)) → U14_aag(T77, T80, X128, T79, appendc31_in_aag(T80, X128, T79))
U14_aag(T77, T80, X128, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The argument filtering Pi contains the following mapping:
fl1_in_aga(x1, x2, x3)  =  fl1_in_aga(x2)
[]  =  []
p6_in_aaaa(x1, x2, x3, x4)  =  p6_in_aaaa
.(x1, x2)  =  .(x1, x2)
append31_in_aag(x1, x2, x3)  =  append31_in_aag(x3)
appendc31_in_aag(x1, x2, x3)  =  appendc31_in_aag(x3)
appendc31_out_aag(x1, x2, x3)  =  appendc31_out_aag(x1, x2, x3)
U14_aag(x1, x2, x3, x4, x5)  =  U14_aag(x1, x4, x5)
s(x1)  =  s(x1)
FL1_IN_AGA(x1, x2, x3)  =  FL1_IN_AGA(x2)
U3_AGA(x1, x2, x3, x4)  =  U3_AGA(x4)
P6_IN_AAAA(x1, x2, x3, x4)  =  P6_IN_AAAA
U1_AAAA(x1, x2, x3, x4)  =  U1_AAAA(x4)
U4_AGA(x1, x2, x3, x4)  =  U4_AGA(x2, x4)
U5_AGA(x1, x2, x3, x4, x5, x6)  =  U5_AGA(x1, x4, x6)
APPEND31_IN_AAG(x1, x2, x3)  =  APPEND31_IN_AAG(x3)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x4, x5)
U6_AGA(x1, x2, x3, x4, x5, x6)  =  U6_AGA(x1, x4, x6)
U7_AGA(x1, x2, x3, x4, x5, x6)  =  U7_AGA(x1, x4, x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APPEND31_IN_AAG(.(T77, T80), X128, .(T77, T79)) → APPEND31_IN_AAG(T80, X128, T79)

The TRS R consists of the following rules:

appendc31_in_aag([], T70, T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T80), X128, .(T77, T79)) → U14_aag(T77, T80, X128, T79, appendc31_in_aag(T80, X128, T79))
U14_aag(T77, T80, X128, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
appendc31_in_aag(x1, x2, x3)  =  appendc31_in_aag(x3)
appendc31_out_aag(x1, x2, x3)  =  appendc31_out_aag(x1, x2, x3)
U14_aag(x1, x2, x3, x4, x5)  =  U14_aag(x1, x4, x5)
APPEND31_IN_AAG(x1, x2, x3)  =  APPEND31_IN_AAG(x3)

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:

APPEND31_IN_AAG(.(T77, T80), X128, .(T77, T79)) → APPEND31_IN_AAG(T80, X128, T79)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND31_IN_AAG(x1, x2, x3)  =  APPEND31_IN_AAG(x3)

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:

APPEND31_IN_AAG(.(T77, T79)) → APPEND31_IN_AAG(T79)

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:

  • APPEND31_IN_AAG(.(T77, T79)) → APPEND31_IN_AAG(T79)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

P6_IN_AAAA([], [], .(T24, T25), s(T26)) → P6_IN_AAAA(T24, X38, T25, T26)

The TRS R consists of the following rules:

appendc31_in_aag([], T70, T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T80), X128, .(T77, T79)) → U14_aag(T77, T80, X128, T79, appendc31_in_aag(T80, X128, T79))
U14_aag(T77, T80, X128, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
appendc31_in_aag(x1, x2, x3)  =  appendc31_in_aag(x3)
appendc31_out_aag(x1, x2, x3)  =  appendc31_out_aag(x1, x2, x3)
U14_aag(x1, x2, x3, x4, x5)  =  U14_aag(x1, x4, x5)
s(x1)  =  s(x1)
P6_IN_AAAA(x1, x2, x3, x4)  =  P6_IN_AAAA

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:

P6_IN_AAAA([], [], .(T24, T25), s(T26)) → P6_IN_AAAA(T24, X38, T25, T26)

R is empty.
The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
P6_IN_AAAA(x1, x2, x3, x4)  =  P6_IN_AAAA

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:

P6_IN_AAAA → P6_IN_AAAA

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

(19) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = P6_IN_AAAA evaluates to t =P6_IN_AAAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P6_IN_AAAA to P6_IN_AAAA.



(20) NO

(21) Obligation:

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

FL1_IN_AGA(.(.(T53, T56), T62), .(T53, T55), s(T63)) → U6_AGA(T53, T56, T62, T55, T63, appendc31_in_aag(T56, T61, T55))
U6_AGA(T53, T56, T62, T55, T63, appendc31_out_aag(T56, T61, T55)) → FL1_IN_AGA(T62, T61, T63)
FL1_IN_AGA(.([], T45), T44, s(T46)) → FL1_IN_AGA(T45, T44, T46)

The TRS R consists of the following rules:

appendc31_in_aag([], T70, T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T80), X128, .(T77, T79)) → U14_aag(T77, T80, X128, T79, appendc31_in_aag(T80, X128, T79))
U14_aag(T77, T80, X128, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
appendc31_in_aag(x1, x2, x3)  =  appendc31_in_aag(x3)
appendc31_out_aag(x1, x2, x3)  =  appendc31_out_aag(x1, x2, x3)
U14_aag(x1, x2, x3, x4, x5)  =  U14_aag(x1, x4, x5)
s(x1)  =  s(x1)
FL1_IN_AGA(x1, x2, x3)  =  FL1_IN_AGA(x2)
U6_AGA(x1, x2, x3, x4, x5, x6)  =  U6_AGA(x1, x4, x6)

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

(22) PiDPToQDPProof (SOUND transformation)

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

(23) Obligation:

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

FL1_IN_AGA(.(T53, T55)) → U6_AGA(T53, T55, appendc31_in_aag(T55))
U6_AGA(T53, T55, appendc31_out_aag(T56, T61, T55)) → FL1_IN_AGA(T61)
FL1_IN_AGA(T44) → FL1_IN_AGA(T44)

The TRS R consists of the following rules:

appendc31_in_aag(T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T79)) → U14_aag(T77, T79, appendc31_in_aag(T79))
U14_aag(T77, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The set Q consists of the following terms:

appendc31_in_aag(x0)
U14_aag(x0, x1, x2)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FL1_IN_AGA(.(T53, T55)) → U6_AGA(T53, T55, appendc31_in_aag(T55))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(.(x1, x2)) = 1 + x1 + x2   
POL(FL1_IN_AGA(x1)) = x1   
POL(U14_aag(x1, x2, x3)) = x1 + x3   
POL(U6_AGA(x1, x2, x3)) = x1 + x3   
POL([]) = 0   
POL(appendc31_in_aag(x1)) = x1   
POL(appendc31_out_aag(x1, x2, x3)) = x2   

The following usable rules [FROCOS05] were oriented:

appendc31_in_aag(T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T79)) → U14_aag(T77, T79, appendc31_in_aag(T79))
U14_aag(T77, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

(25) Obligation:

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

U6_AGA(T53, T55, appendc31_out_aag(T56, T61, T55)) → FL1_IN_AGA(T61)
FL1_IN_AGA(T44) → FL1_IN_AGA(T44)

The TRS R consists of the following rules:

appendc31_in_aag(T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T79)) → U14_aag(T77, T79, appendc31_in_aag(T79))
U14_aag(T77, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The set Q consists of the following terms:

appendc31_in_aag(x0)
U14_aag(x0, x1, x2)

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

(26) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(27) Obligation:

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

FL1_IN_AGA(T44) → FL1_IN_AGA(T44)

The TRS R consists of the following rules:

appendc31_in_aag(T70) → appendc31_out_aag([], T70, T70)
appendc31_in_aag(.(T77, T79)) → U14_aag(T77, T79, appendc31_in_aag(T79))
U14_aag(T77, T79, appendc31_out_aag(T80, X128, T79)) → appendc31_out_aag(.(T77, T80), X128, .(T77, T79))

The set Q consists of the following terms:

appendc31_in_aag(x0)
U14_aag(x0, x1, x2)

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

(28) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(29) Obligation:

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

FL1_IN_AGA(T44) → FL1_IN_AGA(T44)

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

appendc31_in_aag(x0)
U14_aag(x0, x1, x2)

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

(30) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

appendc31_in_aag(x0)
U14_aag(x0, x1, x2)

(31) Obligation:

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

FL1_IN_AGA(T44) → FL1_IN_AGA(T44)

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

(32) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = FL1_IN_AGA(T44) evaluates to t =FL1_IN_AGA(T44)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FL1_IN_AGA(T44) to FL1_IN_AGA(T44).



(33) NO