:: BVFUNC14 semantic presentation

theorem Th1: :: BVFUNC14:1
for b1 being non empty set
for b2 being Element of b1
for b3, b4 being a_partition of b1 holds EqClass b2,(b3 '/\' b4) = (EqClass b2,b3) /\ (EqClass b2,b4)
proof end;

theorem Th2: :: BVFUNC14:2
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4 being a_partition of b1 holds
( b2 = {b3,b4} & b3 <> b4 implies '/\' b2 = b3 '/\' b4 )
proof end;

Lemma2: for b1, b2, b3, b4, b5, b6 being set holds dom (((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6)) = {b1,b2,b3}
proof end;

Lemma3: for b1 being Function
for b2, b3, b4, b5 being set holds
( b2 <> b3 implies ((b1 +* (b2 .--> b4)) +* (b3 .--> b5)) . b2 = b4 )
proof end;

Lemma4: for b1, b2, b3, b4, b5, b6 being set holds
( b1 <> b2 & b3 <> b1 implies (((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6)) . b1 = b4 )
proof end;

Lemma5: for b1, b2, b3, b4, b5, b6 being set
for b7 being Function holds
( b7 = ((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6) implies rng b7 = {(b7 . b1),(b7 . b2),(b7 . b3)} )
proof end;

theorem Th3: :: BVFUNC14:3
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5 being a_partition of b1 holds
( b2 = {b3,b4,b5} & b3 <> b4 & b4 <> b5 & b5 <> b3 implies '/\' b2 = (b3 '/\' b4) '/\' b5 )
proof end;

theorem Th4: :: BVFUNC14:4
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5 being a_partition of b1 holds
( b2 = {b3,b4,b5} & b3 <> b4 & b5 <> b3 implies CompF b3,b2 = b4 '/\' b5 )
proof end;

theorem Th5: :: BVFUNC14:5
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5 being a_partition of b1 holds
( b2 = {b3,b4,b5} & b3 <> b4 & b4 <> b5 implies CompF b4,b2 = b5 '/\' b3 )
proof end;

theorem Th6: :: BVFUNC14:6
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5 being a_partition of b1 holds
( b2 = {b3,b4,b5} & b4 <> b5 & b5 <> b3 implies CompF b5,b2 = b3 '/\' b4 )
proof end;

theorem Th7: :: BVFUNC14:7
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1 holds
( b2 = {b3,b4,b5,b6} & b3 <> b4 & b3 <> b5 & b3 <> b6 implies CompF b3,b2 = (b4 '/\' b5) '/\' b6 )
proof end;

theorem Th8: :: BVFUNC14:8
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1 holds
( b2 = {b3,b4,b5,b6} & b3 <> b4 & b4 <> b5 & b4 <> b6 implies CompF b4,b2 = (b3 '/\' b5) '/\' b6 )
proof end;

theorem Th9: :: BVFUNC14:9
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1 holds
( b2 = {b3,b4,b5,b6} & b3 <> b5 & b4 <> b5 & b5 <> b6 implies CompF b5,b2 = (b3 '/\' b4) '/\' b6 )
proof end;

theorem Th10: :: BVFUNC14:10
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1 holds
( b2 = {b3,b4,b5,b6} & b3 <> b6 & b4 <> b6 & b5 <> b6 implies CompF b6,b2 = (b3 '/\' b5) '/\' b4 )
proof end;

theorem Th11: :: BVFUNC14:11
canceled;

theorem Th12: :: BVFUNC14:12
canceled;

theorem Th13: :: BVFUNC14:13
canceled;

theorem Th14: :: BVFUNC14:14
for b1, b2, b3, b4, b5, b6 being set holds dom (((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6)) = {b1,b2,b3} by Lemma2;

theorem Th15: :: BVFUNC14:15
for b1 being Function
for b2, b3, b4, b5 being set holds
( b2 <> b3 implies ((b1 +* (b2 .--> b4)) +* (b3 .--> b5)) . b2 = b4 ) by Lemma3;

theorem Th16: :: BVFUNC14:16
for b1, b2, b3, b4, b5, b6 being set holds
( b1 <> b2 & b3 <> b1 implies (((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6)) . b1 = b4 ) by Lemma4;

theorem Th17: :: BVFUNC14:17
for b1, b2, b3, b4, b5, b6 being set
for b7 being Function holds
( b7 = ((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6) implies rng b7 = {(b7 . b1),(b7 . b2),(b7 . b3)} ) by Lemma5;

theorem Th18: :: BVFUNC14:18
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1
for b7 being Function
for b8, b9, b10, b11 being set holds
( b2 = {b3,b4,b5,b6} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b4 <> b5 & b4 <> b6 & b5 <> b6 & b7 = (((b4 .--> b9) +* (b5 .--> b10)) +* (b6 .--> b11)) +* (b3 .--> b8) implies ( b7 . b4 = b9 & b7 . b5 = b10 & b7 . b6 = b11 ) )
proof end;

theorem Th19: :: BVFUNC14:19
for b1, b2, b3, b4 being set
for b5 being Function
for b6, b7, b8, b9 being set holds
( b5 = (((b2 .--> b7) +* (b3 .--> b8)) +* (b4 .--> b9)) +* (b1 .--> b6) implies dom b5 = {b1,b2,b3,b4} )
proof end;

theorem Th20: :: BVFUNC14:20
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1
for b7 being Function
for b8, b9, b10, b11 being set holds
( b2 = {b3,b4,b5,b6} & b7 = (((b4 .--> b9) +* (b5 .--> b10)) +* (b6 .--> b11)) +* (b3 .--> b8) implies rng b7 = {(b7 . b3),(b7 . b4),(b7 . b5),(b7 . b6)} )
proof end;

theorem Th21: :: BVFUNC14:21
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1
for b7, b8 being Element of b1
for b9 being Function holds
not ( b2 is independent & b2 = {b3,b4,b5,b6} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b4 <> b5 & b4 <> b6 & b5 <> b6 & not EqClass b8,((b4 '/\' b5) '/\' b6) meets EqClass b7,b3 )
proof end;

theorem Th22: :: BVFUNC14:22
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6 being a_partition of b1
for b7, b8 being Element of b1 holds
not ( b2 is independent & b2 = {b3,b4,b5,b6} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b4 <> b5 & b4 <> b6 & b5 <> b6 & EqClass b7,(b5 '/\' b6) = EqClass b8,(b5 '/\' b6) & not EqClass b8,(CompF b3,b2) meets EqClass b7,(CompF b4,b2) )
proof end;

theorem Th23: :: BVFUNC14:23
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5 being a_partition of b1
for b6, b7 being Element of b1 holds
not ( b2 is independent & b2 = {b3,b4,b5} & b3 <> b4 & b4 <> b5 & b5 <> b3 & EqClass b6,b5 = EqClass b7,b5 & not EqClass b7,(CompF b3,b2) meets EqClass b6,(CompF b4,b2) )
proof end;

theorem Th24: :: BVFUNC14:24
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1 holds
( b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 implies CompF b3,b2 = ((b4 '/\' b5) '/\' b6) '/\' b7 )
proof end;

theorem Th25: :: BVFUNC14:25
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b5 <> b6 & b5 <> b7 & b6 <> b7 implies CompF b4,b2 = ((b3 '/\' b5) '/\' b6) '/\' b7 )
proof end;

theorem Th26: :: BVFUNC14:26
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b5 <> b6 & b5 <> b7 & b6 <> b7 implies CompF b5,b2 = ((b3 '/\' b4) '/\' b6) '/\' b7 )
proof end;

theorem Th27: :: BVFUNC14:27
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b5 <> b6 & b5 <> b7 & b6 <> b7 implies CompF b6,b2 = ((b3 '/\' b4) '/\' b5) '/\' b7 )
proof end;

theorem Th28: :: BVFUNC14:28
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b5 <> b6 & b5 <> b7 & b6 <> b7 implies CompF b7,b2 = ((b3 '/\' b4) '/\' b5) '/\' b6 )
proof end;

theorem Th29: :: BVFUNC14:29
for b1, b2, b3, b4, b5 being set
for b6 being Function
for b7, b8, b9, b10, b11 being set holds
( b1 <> b2 & b1 <> b3 & b1 <> b4 & b1 <> b5 & b2 <> b3 & b2 <> b4 & b2 <> b5 & b3 <> b4 & b3 <> b5 & b4 <> b5 & b6 = ((((b2 .--> b8) +* (b3 .--> b9)) +* (b4 .--> b10)) +* (b5 .--> b11)) +* (b1 .--> b7) implies ( b6 . b1 = b7 & b6 . b2 = b8 & b6 . b3 = b9 & b6 . b4 = b10 & b6 . b5 = b11 ) )
proof end;

theorem Th30: :: BVFUNC14:30
for b1, b2, b3, b4, b5 being set
for b6 being Function
for b7, b8, b9, b10, b11 being set holds
( b6 = ((((b2 .--> b8) +* (b3 .--> b9)) +* (b4 .--> b10)) +* (b5 .--> b11)) +* (b1 .--> b7) implies dom b6 = {b1,b2,b3,b4,b5} )
proof end;

theorem Th31: :: BVFUNC14:31
for b1, b2, b3, b4, b5 being set
for b6 being Function
for b7, b8, b9, b10, b11 being set holds
( b6 = ((((b2 .--> b8) +* (b3 .--> b9)) +* (b4 .--> b10)) +* (b5 .--> b11)) +* (b1 .--> b7) implies rng b6 = {(b6 . b1),(b6 . b2),(b6 . b3),(b6 . b4),(b6 . b5)} )
proof end;

theorem Th32: :: BVFUNC14:32
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1
for b8, b9 being Element of b1
for b10 being Function holds
not ( b2 is independent & b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b5 <> b6 & b5 <> b7 & b6 <> b7 & not EqClass b9,(((b4 '/\' b5) '/\' b6) '/\' b7) meets EqClass b8,b3 )
proof end;

theorem Th33: :: BVFUNC14:33
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7 being a_partition of b1
for b8, b9 being Element of b1 holds
not ( b2 is independent & b2 = {b3,b4,b5,b6,b7} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b5 <> b6 & b5 <> b7 & b6 <> b7 & EqClass b8,((b5 '/\' b6) '/\' b7) = EqClass b9,((b5 '/\' b6) '/\' b7) & not EqClass b9,(CompF b3,b2) meets EqClass b8,(CompF b4,b2) )
proof end;

theorem Th34: :: BVFUNC14:34
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1 holds
( b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 implies CompF b3,b2 = (((b4 '/\' b5) '/\' b6) '/\' b7) '/\' b8 )
proof end;

theorem Th35: :: BVFUNC14:35
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 implies CompF b4,b2 = (((b3 '/\' b5) '/\' b6) '/\' b7) '/\' b8 )
proof end;

theorem Th36: :: BVFUNC14:36
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 implies CompF b5,b2 = (((b3 '/\' b4) '/\' b6) '/\' b7) '/\' b8 )
proof end;

theorem Th37: :: BVFUNC14:37
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 implies CompF b6,b2 = (((b3 '/\' b4) '/\' b5) '/\' b7) '/\' b8 )
proof end;

theorem Th38: :: BVFUNC14:38
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 implies CompF b7,b2 = (((b3 '/\' b4) '/\' b5) '/\' b6) '/\' b8 )
proof end;

theorem Th39: :: BVFUNC14:39
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1 holds
( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 implies CompF b8,b2 = (((b3 '/\' b4) '/\' b5) '/\' b6) '/\' b7 )
proof end;

theorem Th40: :: BVFUNC14:40
for b1, b2, b3, b4, b5, b6 being set
for b7 being Function
for b8, b9, b10, b11, b12, b13 being set holds
( b1 <> b2 & b1 <> b3 & b1 <> b4 & b1 <> b5 & b1 <> b6 & b2 <> b3 & b2 <> b4 & b2 <> b5 & b2 <> b6 & b3 <> b4 & b3 <> b5 & b3 <> b6 & b4 <> b5 & b4 <> b6 & b5 <> b6 & b7 = (((((b2 .--> b9) +* (b3 .--> b10)) +* (b4 .--> b11)) +* (b5 .--> b12)) +* (b6 .--> b13)) +* (b1 .--> b8) implies ( b7 . b1 = b8 & b7 . b2 = b9 & b7 . b3 = b10 & b7 . b4 = b11 & b7 . b5 = b12 & b7 . b6 = b13 ) )
proof end;

theorem Th41: :: BVFUNC14:41
for b1, b2, b3, b4, b5, b6 being set
for b7 being Function
for b8, b9, b10, b11, b12, b13 being set holds
( b7 = (((((b2 .--> b9) +* (b3 .--> b10)) +* (b4 .--> b11)) +* (b5 .--> b12)) +* (b6 .--> b13)) +* (b1 .--> b8) implies dom b7 = {b1,b2,b3,b4,b5,b6} )
proof end;

theorem Th42: :: BVFUNC14:42
for b1, b2, b3, b4, b5, b6 being set
for b7 being Function
for b8, b9, b10, b11, b12, b13 being set holds
( b1 <> b2 & b1 <> b3 & b1 <> b4 & b1 <> b5 & b1 <> b6 & b2 <> b3 & b2 <> b4 & b2 <> b5 & b2 <> b6 & b3 <> b4 & b3 <> b5 & b3 <> b6 & b4 <> b5 & b4 <> b6 & b5 <> b6 & b7 = (((((b2 .--> b9) +* (b3 .--> b10)) +* (b4 .--> b11)) +* (b5 .--> b12)) +* (b6 .--> b13)) +* (b1 .--> b8) implies rng b7 = {(b7 . b1),(b7 . b2),(b7 . b3),(b7 . b4),(b7 . b5),(b7 . b6)} )
proof end;

theorem Th43: :: BVFUNC14:43
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1
for b9, b10 being Element of b1
for b11 being Function holds
not ( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 & not EqClass b10,((((b4 '/\' b5) '/\' b6) '/\' b7) '/\' b8) meets EqClass b9,b3 )
proof end;

theorem Th44: :: BVFUNC14:44
for b1 being non empty set
for b2 being Subset of (PARTITIONS b1)
for b3, b4, b5, b6, b7, b8 being a_partition of b1
for b9, b10 being Element of b1
for b11 being Function holds
not ( b2 is independent & b2 = {b3,b4,b5,b6,b7,b8} & b3 <> b4 & b3 <> b5 & b3 <> b6 & b3 <> b7 & b3 <> b8 & b4 <> b5 & b4 <> b6 & b4 <> b7 & b4 <> b8 & b5 <> b6 & b5 <> b7 & b5 <> b8 & b6 <> b7 & b6 <> b8 & b7 <> b8 & EqClass b9,(((b5 '/\' b6) '/\' b7) '/\' b8) = EqClass b10,(((b5 '/\' b6) '/\' b7) '/\' b8) & not EqClass b10,(CompF b3,b2) meets EqClass b9,(CompF b4,b2) )
proof end;