Theorem comanbn 873

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

Assertion

Ref	Expression
comanbn	(a⊥ ∩ b⊥ ) C ((a ≡ c) ∩ (b ≡ c))

Proof of Theorem comanbn

Step	Hyp	Ref	Expression
1		comanb 872	. 2 (a⊥ ∩ b⊥ ) C ((a⊥ ≡ c⊥ ) ∩ (b⊥ ≡ c⊥ ))
2		conb 122	. . . 4 (a ≡ c) = (a⊥ ≡ c⊥ )
3		conb 122	. . . 4 (b ≡ c) = (b⊥ ≡ c⊥ )
4	2, 3	2an 79	. . 3 ((a ≡ c) ∩ (b ≡ c)) = ((a⊥ ≡ c⊥ ) ∩ (b⊥ ≡ c⊥ ))
5	4	ax-r1 35	. 2 ((a⊥ ≡ c⊥ ) ∩ (b⊥ ≡ c⊥ )) = ((a ≡ c) ∩ (b ≡ c))
6	1, 5	cbtr 182	1 (a⊥ ∩ b⊥ ) C ((a ≡ c) ∩ (b ≡ c))

This proof depends on syntax axioms:   C wc 3  ^⊥ wn 4   ≡ tb 5   ∩ wa 7

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: (None)