:: NAT_4 semantic presentation
theorem Th1: :: NAT_4:1
theorem Th2: :: NAT_4:2
theorem Th3: :: NAT_4:3
theorem Th4: :: NAT_4:4
theorem Th5: :: NAT_4:5
theorem Th6: :: NAT_4:6
theorem Th7: :: NAT_4:7
for
b1,
b2 being
Nat st
b2 > 0 &
b1 divides b2 &
b1 <> 1 &
b1 <> b2 holds
( 1
< b1 &
b1 < b2 )
theorem Th8: :: NAT_4:8
theorem Th9: :: NAT_4:9
for
b1,
b2,
b3,
b4 being
Nat st
b1 = (b2 * b3) + b4 &
b4 < b2 & 0
< b4 holds
not
b2 divides b1
theorem Th10: :: NAT_4:10
for
b1,
b2,
b3 being
Nat st
b2 hcf b3 = 1 &
b2 <> 0 &
b3 <> 0 holds
(b2 |^ b1) hcf b3 = 1
theorem Th11: :: NAT_4:11
theorem Th12: :: NAT_4:12
for
b1 being
Nat holds
( (
b1 <= 1 or ex
b2 being
Nat st
(
b2 divides b1 & 1
< b2 &
b2 < b1 ) ) iff not
b1 is
prime )
theorem Th13: :: NAT_4:13
Lemma14:
for b1 being Nat holds
( ( b1 <= 1 or ex b2 being Nat st
( b2 divides b1 & 1 < b2 & b2 * b2 <= b1 & b2 is prime ) ) iff not b1 is prime )
theorem Th14: :: NAT_4:14
for
b1 being
Nat holds
(
b1 is
prime iff (
b1 > 1 & ( for
b2 being
Nat st 1
< b2 &
b2 * b2 <= b1 &
b2 is
prime holds
not
b2 divides b1 ) ) )
theorem Th15: :: NAT_4:15
theorem Th16: :: NAT_4:16
theorem Th17: :: NAT_4:17
theorem Th18: :: NAT_4:18
theorem Th19: :: NAT_4:19
theorem Th20: :: NAT_4:20
theorem Th21: :: NAT_4:21
theorem Th22: :: NAT_4:22
theorem Th23: :: NAT_4:23
theorem Th24: :: NAT_4:24
theorem Th25: :: NAT_4:25
Lemma27:
for b1 being Nat st 1 < b1 & b1 < 5 & b1 is prime & not b1 = 2 holds
b1 = 3
Lemma28:
for b1 being Nat st b1 < 25 holds
for b2 being Nat st 1 < b2 & b2 * b2 <= b1 & b2 is prime & not b2 = 2 holds
b2 = 3
theorem Th26: :: NAT_4:26
theorem Th27: :: NAT_4:27
theorem Th28: :: NAT_4:28
Lemma31:
( not 6 is prime & not 8 is prime & not 9 is prime & not 10 is prime & not 12 is prime & not 14 is prime & not 15 is prime & not 16 is prime & not 18 is prime & not 20 is prime & not 21 is prime & not 22 is prime & not 24 is prime & not 25 is prime & not 26 is prime & not 27 is prime & not 28 is prime )
theorem Th29: :: NAT_4:29
theorem Th30: :: NAT_4:30
Lemma33:
for b1 being Nat st 1 < b1 & b1 < 29 & b1 is prime & not b1 = 2 & not b1 = 3 & not b1 = 5 & not b1 = 7 & not b1 = 11 & not b1 = 13 & not b1 = 17 & not b1 = 19 holds
b1 = 23
Lemma34:
for b1 being Nat st b1 < 841 holds
for b2 being Nat st 1 < b2 & b2 * b2 <= b1 & b2 is prime & not b2 = 2 & not b2 = 3 & not b2 = 5 & not b2 = 7 & not b2 = 11 & not b2 = 13 & not b2 = 17 & not b2 = 19 holds
b2 = 23
theorem Th31: :: NAT_4:31
theorem Th32: :: NAT_4:32
theorem Th33: :: NAT_4:33
theorem Th34: :: NAT_4:34
theorem Th35: :: NAT_4:35
theorem Th36: :: NAT_4:36
theorem Th37: :: NAT_4:37
theorem Th38: :: NAT_4:38
theorem Th39: :: NAT_4:39
theorem Th40: :: NAT_4:40
Lemma45:
for b1 being Nat st 1 <= b1 & b1 < 4001 holds
ex b2 being Prime st
( b1 < b2 & b2 <= 2 * b1 )
theorem Th41: :: NAT_4:41
theorem Th42: :: NAT_4:42
theorem Th43: :: NAT_4:43
theorem Th44: :: NAT_4:44
theorem Th45: :: NAT_4:45
theorem Th46: :: NAT_4:46
theorem Th47: :: NAT_4:47
theorem Th48: :: NAT_4:48
theorem Th49: :: NAT_4:49
Lemma55:
for b1, b2 being Nat
for b3 being Prime
for b4 being FinSequence of REAL st b1 >= 2 & b3 |^ b2 <= 2 * b1 & 2 * b1 < b3 |^ (b2 + 1) & len b4 = 2 * b1 & ( for b5 being Nat st b5 in dom b4 holds
b4 . b5 = [\((2 * b1) / (b3 |^ b5))/] - (2 * [\(b1 / (b3 |^ b5))/]) ) holds
Sum b4 <= b2
Lemma56:
for b1 being Nat
for b2 being Prime st b1 >= 3 holds
( ( b2 > 2 * b1 implies b2 |-count ((2 * b1) choose b1) = 0 ) & ( b1 < b2 & b2 <= 2 * b1 implies b2 |-count ((2 * b1) choose b1) <= 1 ) & ( (2 * b1) / 3 < b2 & b2 <= b1 implies b2 |-count ((2 * b1) choose b1) = 0 ) & ( sqrt (2 * b1) < b2 & b2 <= (2 * b1) / 3 implies b2 |-count ((2 * b1) choose b1) <= 1 ) & ( b2 <= sqrt (2 * b1) implies b2 |^ (b2 |-count ((2 * b1) choose b1)) <= 2 * b1 ) )
:: deftheorem Def1 defines |-count NAT_4:def 1 :
theorem Th50: :: NAT_4:50
theorem Th51: :: NAT_4:51
theorem Th52: :: NAT_4:52
theorem Th53: :: NAT_4:53
theorem Th54: :: NAT_4:54
theorem Th55: :: NAT_4:55
Lemma64:
for b1, b2 being Nat st b2 = (2 * b1) choose b1 & b1 >= 3 holds
b2 = ((Product (Sgm { (b3 |^ (b3 |-count b2)) where B is Prime : ( b3 <= sqrt (2 * b1) & b3 |-count b2 > 0 ) } )) * (Product (Sgm { (b3 |^ (b3 |-count b2)) where B is Prime : ( sqrt (2 * b1) < b3 & b3 <= (2 * b1) / 3 & b3 |-count b2 > 0 ) } ))) * (Product (Sgm { (b3 |^ (b3 |-count b2)) where B is Prime : ( b1 < b3 & b3 <= 2 * b1 & b3 |-count b2 > 0 ) } ))
Lemma65:
for b1, b2 being Nat st b2 = (2 * b1) choose b1 & b1 >= 3 holds
Product (Sgm { (b3 |^ (b3 |-count b2)) where B is Prime : ( b3 <= sqrt (2 * b1) & b3 |-count b2 > 0 ) } ) <= (2 * b1) to_power (sqrt (2 * b1))
theorem Th56: :: NAT_4:56
for
b1 being
Nat st
b1 >= 1 holds
ex
b2 being
Prime st
(
b1 < b2 &
b2 <= 2
* b1 )