:: BVFUNC14 semantic presentation
theorem Th1: :: BVFUNC14:1
theorem Th2: :: BVFUNC14:2
Lemma2:
for b1, b2, b3, b4, b5, b6 being set holds dom (((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6)) = {b1,b2,b3}
Lemma3:
for b1 being Function
for b2, b3, b4, b5 being set st b2 <> b3 holds
((b1 +* (b2 .--> b4)) +* (b3 .--> b5)) . b2 = b4
Lemma4:
for b1, b2, b3, b4, b5, b6 being set st b1 <> b2 & b3 <> b1 holds
(((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6)) . b1 = b4
Lemma5:
for b1, b2, b3, b4, b5, b6 being set
for b7 being Function st b7 = ((b1 .--> b4) +* (b2 .--> b5)) +* (b3 .--> b6) holds
rng b7 = {(b7 . b1),(b7 . b2),(b7 . b3)}
theorem Th3: :: BVFUNC14:3
theorem Th4: :: BVFUNC14:4
theorem Th5: :: BVFUNC14:5
theorem Th6: :: BVFUNC14:6
theorem Th7: :: BVFUNC14:7
theorem Th8: :: BVFUNC14:8
theorem Th9: :: BVFUNC14:9
theorem Th10: :: BVFUNC14:10
theorem Th11: :: BVFUNC14:11
canceled;
theorem Th12: :: BVFUNC14:12
canceled;
theorem Th13: :: BVFUNC14:13
canceled;
theorem Th14: :: BVFUNC14:14
theorem Th15: :: BVFUNC14:15
theorem Th16: :: BVFUNC14:16
theorem Th17: :: BVFUNC14:17
theorem Th18: :: BVFUNC14:18
theorem Th19: :: BVFUNC14:19
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 st
b2 = {b3,b4,b5,b6} &
b7 = (((b4 .--> b9) +* (b5 .--> b10)) +* (b6 .--> b11)) +* (b3 .--> b8) holds
rng b7 = {(b7 . b3),(b7 . b4),(b7 . b5),(b7 . b6)}
theorem Th21: :: BVFUNC14:21
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 st
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) holds
EqClass b8,
(CompF b3,b2) meets EqClass b7,
(CompF b4,b2)
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 st
b2 is
independent &
b2 = {b3,b4,b5} &
b3 <> b4 &
b4 <> b5 &
b5 <> b3 &
EqClass b6,
b5 = EqClass b7,
b5 holds
EqClass b7,
(CompF b3,b2) meets EqClass b6,
(CompF b4,b2)
theorem Th24: :: BVFUNC14:24
theorem Th25: :: BVFUNC14:25
theorem Th26: :: BVFUNC14:26
theorem Th27: :: BVFUNC14:27
theorem Th28: :: BVFUNC14:28
theorem Th29: :: BVFUNC14:29
for
b1,
b2,
b3,
b4,
b5 being
set for
b6 being
Function for
b7,
b8,
b9,
b10,
b11 being
set st
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) holds
(
b6 . b1 = b7 &
b6 . b2 = b8 &
b6 . b3 = b9 &
b6 . b4 = b10 &
b6 . b5 = b11 )
theorem Th30: :: BVFUNC14:30
for
b1,
b2,
b3,
b4,
b5 being
set for
b6 being
Function for
b7,
b8,
b9,
b10,
b11 being
set st
b6 = ((((b2 .--> b8) +* (b3 .--> b9)) +* (b4 .--> b10)) +* (b5 .--> b11)) +* (b1 .--> b7) holds
dom b6 = {b1,b2,b3,b4,b5}
theorem Th31: :: BVFUNC14:31
for
b1,
b2,
b3,
b4,
b5 being
set for
b6 being
Function for
b7,
b8,
b9,
b10,
b11 being
set st
b6 = ((((b2 .--> b8) +* (b3 .--> b9)) +* (b4 .--> b10)) +* (b5 .--> b11)) +* (b1 .--> b7) holds
rng b6 = {(b6 . b1),(b6 . b2),(b6 . b3),(b6 . b4),(b6 . b5)}
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 st
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 holds
EqClass b9,
(((b4 '/\' b5) '/\' b6) '/\' b7) meets EqClass b8,
b3
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 st
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) holds
EqClass b9,
(CompF b3,b2) meets EqClass b8,
(CompF b4,b2)