Article BCIALG_3, MML version 4.99.1005

:: BCIALG_3:attrnot 1 => BCIALG_3:attr 1
definition
  let a1 be non empty BCIStr_0;
  attr a1 is commutative means
    for b1, b2 being Element of the carrier of a1 holds
    b1 \ (b1 \ b2) = b2 \ (b2 \ b1);
end;

:: BCIALG_3:dfs 1
definiens
  let a1 be non empty BCIStr_0;
To prove
     a1 is commutative
it is sufficient to prove
  thus for b1, b2 being Element of the carrier of a1 holds
    b1 \ (b1 \ b2) = b2 \ (b2 \ b1);

:: BCIALG_3:def 1
theorem
for b1 being non empty BCIStr_0 holds
      b1 is commutative
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ b3) = b3 \ (b3 \ b2);

:: BCIALG_3:funcreg 1
registration
  cluster BCI-EXAMPLE -> commutative;
end;

:: BCIALG_3:exreg 1
registration
  cluster non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0;
end;

:: BCIALG_3:th 1
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ b3) <= b3 \ (b3 \ b2);

:: BCIALG_3:th 2
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
for b2, b3 being Element of the carrier of b1 holds
b2 \ (b2 \ b3) <= b3 & b2 \ (b2 \ b3) <= b2;

:: BCIALG_3:th 3
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ b3 = b2 \ (b3 \ (b3 \ b2));

:: BCIALG_3:th 4
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ b3) = b3 \ (b3 \ (b2 \ (b2 \ b3)));

:: BCIALG_3:th 5
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1
            st b2 <= b3
         holds b2 = b3 \ (b3 \ b2);

:: BCIALG_3:th 6
theorem
for b1 being non empty BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      b2 \ ((0. b1) \ b3) = b2 &
       (b2 \ b4) \ (b2 \ b3) = (b3 \ b4) \ (b3 \ b2);

:: BCIALG_3:th 7
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
   st b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
for b2, b3 being Element of the carrier of b1
      st b2 \ b3 = b2
   holds b3 \ b2 = b3;

:: BCIALG_3:th 8
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
   st b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
for b2, b3, b4 being Element of the carrier of b1
      st b3 <= b4
   holds (b4 \ b2) \ (b4 \ b3) = b3 \ b2;

:: BCIALG_3:th 9
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
   st b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
for b2, b3 being Element of the carrier of b1 holds
   b2 \ b3 = b2
iff
   b3 \ (b3 \ b2) = 0. b1;

:: BCIALG_3:th 10
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
   st b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
for b2, b3 being Element of the carrier of b1 holds
b2 \ (b3 \ (b3 \ b2)) = b2 \ b3 &
 (b2 \ b3) \ ((b2 \ b3) \ b2) = b2 \ b3;

:: BCIALG_3:th 11
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
   st b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0
for b2, b3, b4 being Element of the carrier of b1
      st b2 <= b4
   holds (b4 \ b3) \ ((b4 \ b3) \ (b4 \ b2)) = (b4 \ b3) \ (b2 \ b3);

:: BCIALG_3:attrnot 2 => BCIALG_3:attr 2
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
  let a2 be Element of the carrier of a1;
  attr a2 is being_greatest means
    for b1 being Element of the carrier of a1 holds
       b1 \ a2 = 0. a1;
end;

:: BCIALG_3:dfs 2
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
  let a2 be Element of the carrier of a1;
To prove
     a2 is being_greatest
it is sufficient to prove
  thus for b1 being Element of the carrier of a1 holds
       b1 \ a2 = 0. a1;

:: BCIALG_3:def 2
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0
for b2 being Element of the carrier of b1 holds
      b2 is being_greatest(b1)
   iff
      for b3 being Element of the carrier of b1 holds
         b3 \ b2 = 0. b1;

:: BCIALG_3:attrnot 3 => BCIALG_3:attr 3
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
  let a2 be Element of the carrier of a1;
  attr a2 is being_positive means
    (0. a1) \ a2 = 0. a1;
end;

:: BCIALG_3:dfs 3
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
  let a2 be Element of the carrier of a1;
To prove
     a2 is being_positive
it is sufficient to prove
  thus (0. a1) \ a2 = 0. a1;

:: BCIALG_3:def 3
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0
for b2 being Element of the carrier of b1 holds
      b2 is being_positive(b1)
   iff
      (0. b1) \ b2 = 0. b1;

:: BCIALG_3:attrnot 4 => BCIALG_3:attr 4
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
  attr a1 is BCI-commutative means
    for b1, b2 being Element of the carrier of a1
          st b1 \ b2 = 0. a1
       holds b1 = b2 \ (b2 \ b1);
end;

:: BCIALG_3:dfs 4
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
To prove
     a1 is BCI-commutative
it is sufficient to prove
  thus for b1, b2 being Element of the carrier of a1
          st b1 \ b2 = 0. a1
       holds b1 = b2 \ (b2 \ b1);

:: BCIALG_3:def 4
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-commutative
   iff
      for b2, b3 being Element of the carrier of b1
            st b2 \ b3 = 0. b1
         holds b2 = b3 \ (b3 \ b2);

:: BCIALG_3:attrnot 5 => BCIALG_3:attr 5
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
  attr a1 is BCI-weakly-commutative means
    for b1, b2 being Element of the carrier of a1 holds
    (b1 \ (b1 \ b2)) \ ((0. a1) \ (b1 \ b2)) = b2 \ (b2 \ b1);
end;

:: BCIALG_3:dfs 5
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 BCIStr_0;
To prove
     a1 is BCI-weakly-commutative
it is sufficient to prove
  thus for b1, b2 being Element of the carrier of a1 holds
    (b1 \ (b1 \ b2)) \ ((0. a1) \ (b1 \ b2)) = b2 \ (b2 \ b1);

:: BCIALG_3:def 5
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-weakly-commutative
   iff
      for b2, b3 being Element of the carrier of b1 holds
      (b2 \ (b2 \ b3)) \ ((0. b1) \ (b2 \ b3)) = b3 \ (b3 \ b2);

:: BCIALG_3:funcreg 2
registration
  cluster BCI-EXAMPLE -> BCI-commutative BCI-weakly-commutative;
end;

:: BCIALG_3:exreg 2
registration
  cluster non empty being_B being_C being_I being_BCI-4 BCI-commutative BCI-weakly-commutative BCIStr_0;
end;

:: BCIALG_3:th 12
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0
      st ex b2 being Element of the carrier of b1 st
           b2 is being_greatest(b1)
   holds b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;

:: BCIALG_3:th 13
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0
      st b1 is p-Semisimple
   holds b1 is BCI-commutative & b1 is BCI-weakly-commutative;

:: BCIALG_3:th 14
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0 holds
   b1 is non empty being_B being_C being_I being_BCI-4 BCI-commutative BCIStr_0 &
    b1 is non empty being_B being_C being_I being_BCI-4 BCI-weakly-commutative BCIStr_0;

:: BCIALG_3:th 15
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
      st b1 is non empty being_B being_C being_I being_BCI-4 BCI-weakly-commutative BCIStr_0
   holds b1 is BCI-commutative;

:: BCIALG_3:th 16
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-commutative
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ b3) = b3 \ (b3 \ (b2 \ (b2 \ b3)));

:: BCIALG_3:th 17
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-commutative
   iff
      for b2, b3 being Element of the carrier of b1 holds
      (b2 \ (b2 \ b3)) \ (b3 \ (b3 \ b2)) = (0. b1) \ (b2 \ b3);

:: BCIALG_3:th 18
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-commutative
   iff
      for b2 being Element of AtomSet b1
      for b3, b4 being Element of BranchV b2 holds
      b3 \ (b3 \ b4) = b4 \ (b4 \ b3);

:: BCIALG_3:th 19
theorem
for b1 being non empty BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 BCI-commutative BCIStr_0
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      ((b2 \ b3) \ (b2 \ b4)) \ (b4 \ b3) = 0. b1 &
       b2 \ 0. b1 = b2 &
       b2 \ (b2 \ b3) = b3 \ (b3 \ (b2 \ (b2 \ b3)));

:: BCIALG_3:th 20
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-commutative
   iff
      for b2, b3, b4 being Element of the carrier of b1
            st b2 <= b4 & b4 \ b3 <= b4 \ b2
         holds b2 <= b3;

:: BCIALG_3:th 21
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 BCIStr_0 holds
      b1 is BCI-commutative
   iff
      for b2, b3, b4 being Element of the carrier of b1
            st b2 <= b3 & b2 <= b4
         holds b2 <= b3 \ (b3 \ b4);

:: BCIALG_3:attrnot 6 => BCIALG_3:attr 6
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  attr a1 is bounded means
    ex b1 being Element of the carrier of a1 st
       b1 is being_greatest(a1);
end;

:: BCIALG_3:dfs 6
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
To prove
     a1 is bounded
it is sufficient to prove
  thus ex b1 being Element of the carrier of a1 st
       b1 is being_greatest(a1);

:: BCIALG_3:def 6
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is bounded
   iff
      ex b2 being Element of the carrier of b1 st
         b2 is being_greatest(b1);

:: BCIALG_3:funcreg 3
registration
  cluster BCI-EXAMPLE -> bounded;
end;

:: BCIALG_3:exreg 3
registration
  cluster non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0;
end;

:: BCIALG_3:attrnot 7 => BCIALG_3:attr 7
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0;
  attr a1 is involutory means
    for b1 being Element of the carrier of a1
       st b1 is being_greatest(a1)
    for b2 being Element of the carrier of a1 holds
       b1 \ (b1 \ b2) = b2;
end;

:: BCIALG_3:dfs 7
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0;
To prove
     a1 is involutory
it is sufficient to prove
  thus for b1 being Element of the carrier of a1
       st b1 is being_greatest(a1)
    for b2 being Element of the carrier of a1 holds
       b1 \ (b1 \ b2) = b2;

:: BCIALG_3:def 7
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0 holds
      b1 is involutory
   iff
      for b2 being Element of the carrier of b1
         st b2 is being_greatest(b1)
      for b3 being Element of the carrier of b1 holds
         b2 \ (b2 \ b3) = b3;

:: BCIALG_3:th 22
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0 holds
      b1 is involutory
   iff
      for b2 being Element of the carrier of b1
         st b2 is being_greatest(b1)
      for b3, b4 being Element of the carrier of b1 holds
      b3 \ b4 = (b2 \ b4) \ (b2 \ b3);

:: BCIALG_3:th 23
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0 holds
      b1 is involutory
   iff
      for b2 being Element of the carrier of b1
         st b2 is being_greatest(b1)
      for b3, b4 being Element of the carrier of b1 holds
      b3 \ (b2 \ b4) = b4 \ (b2 \ b3);

:: BCIALG_3:th 24
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0 holds
      b1 is involutory
   iff
      for b2 being Element of the carrier of b1
         st b2 is being_greatest(b1)
      for b3, b4 being Element of the carrier of b1
            st b3 <= b2 \ b4
         holds b4 <= b2 \ b3;

:: BCIALG_3:attrnot 8 => BCIALG_3:attr 8
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  let a2 be Element of the carrier of a1;
  attr a2 is being_Iseki means
    for b1 being Element of the carrier of a1 holds
       b1 \ a2 = 0. a1 & a2 \ b1 = a2;
end;

:: BCIALG_3:dfs 8
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  let a2 be Element of the carrier of a1;
To prove
     a2 is being_Iseki
it is sufficient to prove
  thus for b1 being Element of the carrier of a1 holds
       b1 \ a2 = 0. a1 & a2 \ b1 = a2;

:: BCIALG_3:def 8
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
for b2 being Element of the carrier of b1 holds
      b2 is being_Iseki(b1)
   iff
      for b3 being Element of the carrier of b1 holds
         b3 \ b2 = 0. b1 & b2 \ b3 = b2;

:: BCIALG_3:attrnot 9 => BCIALG_3:attr 9
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  attr a1 is Iseki_extension means
    ex b1 being Element of the carrier of a1 st
       b1 is being_Iseki(a1);
end;

:: BCIALG_3:dfs 9
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
To prove
     a1 is Iseki_extension
it is sufficient to prove
  thus ex b1 being Element of the carrier of a1 st
       b1 is being_Iseki(a1);

:: BCIALG_3:def 9
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is Iseki_extension
   iff
      ex b2 being Element of the carrier of b1 st
         b2 is being_Iseki(b1);

:: BCIALG_3:funcreg 4
registration
  cluster BCI-EXAMPLE -> Iseki_extension;
end;

:: BCIALG_3:modenot 1 => BCIALG_3:mode 1
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  mode Commutative-Ideal of A1 -> non empty Element of bool the carrier of a1 means
    0. a1 in it &
     (for b1, b2, b3 being Element of the carrier of a1
           st (b1 \ b2) \ b3 in it & b3 in it
        holds b1 \ (b2 \ (b2 \ b1)) in it);
end;

:: BCIALG_3:dfs 10
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  let a2 be non empty Element of bool the carrier of a1;
To prove
     a2 is Commutative-Ideal of a1
it is sufficient to prove
  thus 0. a1 in a2 &
     (for b1, b2, b3 being Element of the carrier of a1
           st (b1 \ b2) \ b3 in a2 & b3 in a2
        holds b1 \ (b2 \ (b2 \ b1)) in a2);

:: BCIALG_3:def 10
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
for b2 being non empty Element of bool the carrier of b1 holds
      b2 is Commutative-Ideal of b1
   iff
      0. b1 in b2 &
       (for b3, b4, b5 being Element of the carrier of b1
             st (b3 \ b4) \ b5 in b2 & b5 in b2
          holds b3 \ (b4 \ (b4 \ b3)) in b2);

:: BCIALG_3:th 25
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
for b2 being non empty Element of bool the carrier of b1
   st b2 is Commutative-Ideal of b1
for b3, b4 being Element of the carrier of b1
      st b3 \ b4 in b2
   holds b3 \ (b4 \ (b4 \ b3)) in b2;

:: BCIALG_3:th 26
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
for b2 being non empty Element of bool the carrier of b1
for b3 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
      st b2 is Commutative-Ideal of b3
   holds b2 is Ideal of b3;

:: BCIALG_3:th 27
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0
for b2 being non empty Element of bool the carrier of b1
   st b2 is Commutative-Ideal of b1
for b3, b4 being Element of the carrier of b1
      st b3 \ (b3 \ b4) in b2
   holds (b4 \ (b4 \ b3)) \ (b3 \ b4) in b2;

:: BCIALG_3:attrnot 10 => BCIALG_3:attr 10
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  attr a1 is BCK-positive-implicative means
    for b1, b2, b3 being Element of the carrier of a1 holds
    (b1 \ b2) \ b3 = (b1 \ b3) \ (b2 \ b3);
end;

:: BCIALG_3:dfs 11
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
To prove
     a1 is BCK-positive-implicative
it is sufficient to prove
  thus for b1, b2, b3 being Element of the carrier of a1 holds
    (b1 \ b2) \ b3 = (b1 \ b3) \ (b2 \ b3);

:: BCIALG_3:def 11
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is BCK-positive-implicative
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      (b2 \ b3) \ b4 = (b2 \ b4) \ (b3 \ b4);

:: BCIALG_3:attrnot 11 => BCIALG_3:attr 11
definition
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
  attr a1 is BCK-implicative means
    for b1, b2 being Element of the carrier of a1 holds
    b1 \ (b2 \ b1) = b1;
end;

:: BCIALG_3:dfs 12
definiens
  let a1 be non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0;
To prove
     a1 is BCK-implicative
it is sufficient to prove
  thus for b1, b2 being Element of the carrier of a1 holds
    b1 \ (b2 \ b1) = b1;

:: BCIALG_3:def 12
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is BCK-implicative
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b3 \ b2) = b2;

:: BCIALG_3:funcreg 5
registration
  cluster BCI-EXAMPLE -> BCK-positive-implicative BCK-implicative;
end;

:: BCIALG_3:exreg 4
registration
  cluster non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded Iseki_extension BCK-positive-implicative BCK-implicative BCIStr_0;
end;

:: BCIALG_3:th 28
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ b3 = (b2 \ b3) \ b3;

:: BCIALG_3:th 29
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      (b2 \ (b2 \ b3)) \ (b3 \ b2) = b2 \ (b2 \ (b3 \ (b3 \ b2)));

:: BCIALG_3:th 30
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ b3 = (b2 \ b3) \ (b2 \ (b2 \ b3));

:: BCIALG_3:th 31
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      (b2 \ b4) \ (b3 \ b4) <= (b2 \ b3) \ b4;

:: BCIALG_3:th 32
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ b3 <= (b2 \ b3) \ b3;

:: BCIALG_3:th 33
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ (b3 \ (b3 \ b2))) <= (b2 \ (b2 \ b3)) \ (b3 \ b2);

:: BCIALG_3:th 34
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative BCIStr_0 &
       b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-positive-implicative BCIStr_0;

:: BCIALG_3:th 35
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      (b2 \ (b2 \ b3)) \ (b2 \ b3) = b3 \ (b3 \ b2);

:: BCIALG_3:th 36
theorem
for b1 being non empty BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      b2 \ ((0. b1) \ b3) = b2 &
       (b2 \ b4) \ (b2 \ b3) = ((b3 \ b4) \ (b3 \ b2)) \ (b2 \ b3);

:: BCIALG_3:th 37
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      b1 is involutory & b1 is BCK-positive-implicative;

:: BCIALG_3:th 38
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ (b3 \ b2)) = 0. b1;

:: BCIALG_3:th 39
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      (b2 \ (b2 \ b3)) \ (b2 \ b3) = b3 \ (b3 \ (b2 \ (b2 \ b3)));

:: BCIALG_3:th 40
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      (b2 \ b4) \ (b2 \ b3) = (b3 \ b4) \ ((b3 \ b2) \ b4);

:: BCIALG_3:th 41
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3, b4 being Element of the carrier of b1 holds
      b2 \ (b2 \ (b3 \ b4)) = (b3 \ b4) \ ((b3 \ b4) \ (b2 \ b4));

:: BCIALG_3:th 42
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCIStr_0 holds
      b1 is non empty being_B being_C being_I being_BCI-4 being_BCK-5 BCK-implicative BCIStr_0
   iff
      for b2, b3 being Element of the carrier of b1 holds
      b2 \ (b2 \ b3) = (b3 \ (b3 \ b2)) \ (b2 \ b3);

:: BCIALG_3:th 43
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      for b3 being Element of the carrier of b1 holds
         (b2 \ b3) \ ((b2 \ b3) \ b3) = 0. b1;

:: BCIALG_3:th 44
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      for b3 being Element of the carrier of b1 holds
         b3 \ (b2 \ b3) = b3;

:: BCIALG_3:th 45
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      for b3 being Element of the carrier of b1 holds
         (b2 \ b3) \ b3 = b2 \ b3;

:: BCIALG_3:th 46
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      for b3, b4 being Element of the carrier of b1 holds
      (b2 \ b4) \ ((b2 \ b4) \ b3) = b3 \ b4;

:: BCIALG_3:th 47
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      for b3, b4 being Element of the carrier of b1 holds
      b4 \ (b4 \ b3) = b3 \ (b2 \ b4);

:: BCIALG_3:th 48
theorem
for b1 being non empty being_B being_C being_I being_BCI-4 being_BCK-5 commutative bounded BCIStr_0
for b2 being Element of the carrier of b1
      st b2 is being_greatest(b1)
   holds    b1 is BCK-implicative
   iff
      for b3, b4, b5 being Element of the carrier of b1 holds
      (b3 \ (b4 \ b5)) \ (b3 \ b4) <= b3 \ (b2 \ b5);