:: FIB_NUM semantic presentation
theorem Th1: :: FIB_NUM:1
for
b1,
b2 being
Nat holds
b1 hcf b2 = b1 hcf (b2 + b1)
theorem Th2: :: FIB_NUM:2
for
b1,
b2,
b3 being
Nat st
b1 hcf b2 = 1 holds
b1 hcf (b2 * b3) = b1 hcf b3
theorem Th3: :: FIB_NUM:3
scheme :: FIB_NUM:sch 2
s2{
P1[
Nat,
Nat] } :
for
b1,
b2 being
Nat holds
P1[
b1,
b2]
provided
E4:
for
b1,
b2 being
Nat st
P1[
b1,
b2] holds
P1[
b2,
b1]
and E5:
for
b1 being
Nat st ( for
b2,
b3 being
Nat st
b2 < b1 &
b3 < b1 holds
P1[
b2,
b3] ) holds
for
b2 being
Nat st
b2 <= b1 holds
P1[
b1,
b2]
(0 + 1) + 1 = 2
;
then Lemma4:
Fib 2 = 1
by PRE_FF:1;
Lemma5:
(1 + 1) + 1 = 3
;
Lemma6:
for b1 being Nat holds Fib (b1 + 1) >= b1
Lemma7:
for b1 being Nat holds Fib (b1 + 1) >= Fib b1
Lemma8:
for b1, b2 being Nat st b1 >= b2 holds
Fib b1 >= Fib b2
Lemma9:
for b1 being Nat holds Fib (b1 + 1) <> 0
theorem Th4: :: FIB_NUM:4
Lemma11:
for b1 being Nat holds (Fib b1) hcf (Fib (b1 + 1)) = 1
theorem Th5: :: FIB_NUM:5
theorem Th6: :: FIB_NUM:6
:: deftheorem Def1 defines tau FIB_NUM:def 1 :
:: deftheorem Def2 defines tau_bar FIB_NUM:def 2 :
Lemma13:
( tau ^2 = tau + 1 & tau_bar ^2 = tau_bar + 1 )
Lemma14:
2 < sqrt 5
by SQUARE_1:85, SQUARE_1:95;
Lemma15:
sqrt 5 <> 0
by SQUARE_1:85, SQUARE_1:95;
Lemma16:
sqrt 5 < 3
1 < tau
then Lemma17:
0 < tau
by XREAL_1:2;
Lemma18:
tau_bar < 0
Lemma19:
abs tau_bar < 1
theorem Th7: :: FIB_NUM:7
Lemma21:
for b1 being Nat
for b2 being real number st abs b2 <= 1 holds
abs (b2 |^ b1) <= 1
Lemma22:
for b1 being Nat holds abs ((tau_bar to_power b1) / (sqrt 5)) < 1
theorem Th8: :: FIB_NUM:8
theorem Th9: :: FIB_NUM:9
theorem Th10: :: FIB_NUM:10
theorem Th11: :: FIB_NUM:11
theorem Th12: :: FIB_NUM:12