Theorem wnbdi 411

Description: Negated biconditional (distributive form)

Assertion

Ref	Expression
wnbdi	((a ≡ b)⊥ ≡ (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))) = 1

Proof of Theorem wnbdi

Step	Hyp	Ref	Expression
1		dfnb 87	. . 3 (a ≡ b)⊥ = ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))
2	1	bi1 110	. 2 ((a ≡ b)⊥ ≡ ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))) = 1
3		wcomorr 394	. . . . 5 C (a, (a ∪ b)) = 1
4	3	wcomcom 396	. . . 4 C ((a ∪ b), a) = 1
5	4	wcomcom2 397	. . 3 C ((a ∪ b), a⊥ ) = 1
6		wcomorr 394	. . . . . 6 C (b, (b ∪ a)) = 1
7		ax-a2 30	. . . . . . 7 (b ∪ a) = (a ∪ b)
8	7	bi1 110	. . . . . 6 ((b ∪ a) ≡ (a ∪ b)) = 1
9	6, 8	wcbtr 393	. . . . 5 C (b, (a ∪ b)) = 1
10	9	wcomcom 396	. . . 4 C ((a ∪ b), b) = 1
11	10	wcomcom2 397	. . 3 C ((a ∪ b), b⊥ ) = 1
12	5, 11	wfh1 405	. 2 (((a ∪ b) ∩ (a⊥ ∪ b⊥ )) ≡ (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))) = 1
13	2, 12	wr2 353	1 ((a ≡ b)⊥ ≡ (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 9

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-wom 343

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-i2 44  df-le 121  df-le1 122  df-le2 123  df-cmtr 126

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