Theorem comanbn 855

Description: Biconditional commutation law.

Assertion

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

Proof of Theorem comanbn

Step	Hyp	Ref	Expression
1		comanb 854	. 2 (a⊥ ∩ b⊥ ) C ((a⊥ ≡ c⊥ ) ∩ (b⊥ ≡ c⊥ ))
2		conb 114	. . . 4 (a ≡ c) = (a⊥ ≡ c⊥ )
3		conb 114	. . . 4 (b ≡ c) = (b⊥ ≡ c⊥ )
4	2, 3	2an 72	. . 3 ((a ≡ c) ∩ (b ≡ c)) = ((a⊥ ≡ c⊥ ) ∩ (b⊥ ≡ c⊥ ))
5	4	ax-r1 34	. 2 ((a⊥ ≡ c⊥ ) ∩ (b⊥ ≡ c⊥ )) = ((a ≡ c) ∩ (b ≡ c))
6	1, 5	cbtr 174	1 (a⊥ ∩ b⊥ ) C ((a ≡ c) ∩ (b ≡ c))

Syntax hints:   C wc 3  ^⊥ wn 4   ≡ tb 5   ∩ wa 7

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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