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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