Theorem comanb 872

Description: Biconditional commutation law. (Contributed by NM, 1-Dec-1999.)

Assertion

Ref	Expression
comanb	(a ∩ b) C ((a ≡ c) ∩ (b ≡ c))

Proof of Theorem comanb

Step	Hyp	Ref	Expression
1		lea 160	. . . 4 (((a ∪ c)⊥ ∪ ((a ∩ b) ∩ c)) ∩ (b →1 c)) ≤ ((a ∪ c)⊥ ∪ ((a ∩ b) ∩ c))
2		lea 160	. . . . . . 7 (a ∩ b) ≤ a
3		leo 158	. . . . . . 7 a ≤ (a ∪ c)
4	2, 3	letr 137	. . . . . 6 (a ∩ b) ≤ (a ∪ c)
5	4	lecon 154	. . . . 5 (a ∪ c)⊥ ≤ (a ∩ b)⊥
6	5	leror 152	. . . 4 ((a ∪ c)⊥ ∪ ((a ∩ b) ∩ c)) ≤ ((a ∩ b)⊥ ∪ ((a ∩ b) ∩ c))
7	1, 6	letr 137	. . 3 (((a ∪ c)⊥ ∪ ((a ∩ b) ∩ c)) ∩ (b →1 c)) ≤ ((a ∩ b)⊥ ∪ ((a ∩ b) ∩ c))
8		comanblem1 870	. . 3 ((a ≡ c) ∩ (b ≡ c)) = (((a ∪ c)⊥ ∪ ((a ∩ b) ∩ c)) ∩ (b →1 c))
9		df-i1 44	. . . 4 ((a ∩ b) →1 ((a ≡ c) ∩ (b ≡ c))) = ((a ∩ b)⊥ ∪ ((a ∩ b) ∩ ((a ≡ c) ∩ (b ≡ c))))
10		comanblem2 871	. . . . 5 ((a ∩ b) ∩ ((a ≡ c) ∩ (b ≡ c))) = ((a ∩ b) ∩ c)
11	10	lor 70	. . . 4 ((a ∩ b)⊥ ∪ ((a ∩ b) ∩ ((a ≡ c) ∩ (b ≡ c)))) = ((a ∩ b)⊥ ∪ ((a ∩ b) ∩ c))
12	9, 11	ax-r2 36	. . 3 ((a ∩ b) →1 ((a ≡ c) ∩ (b ≡ c))) = ((a ∩ b)⊥ ∪ ((a ∩ b) ∩ c))
13	7, 8, 12	le3tr1 140	. 2 ((a ≡ c) ∩ (b ≡ c)) ≤ ((a ∩ b) →1 ((a ≡ c) ∩ (b ≡ c)))
14	13	i1com 708	1 (a ∩ b) C ((a ≡ c) ∩ (b ≡ c))

This proof depends on syntax axioms:   C wc 3  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  comanbn  873