Theorem wnbdi 429

Description: Negated biconditional (distributive form). (Contributed by NM, 13-Oct-1997.)

Assertion

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

Proof of Theorem wnbdi

Step	Hyp	Ref	Expression
1		dfnb 95	. . 3 (a ≡ b)⊥ = ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))
2	1	bi1 118	. 2 ((a ≡ b)⊥ ≡ ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))) = 1
3		wcomorr 412	. . . . 5 C (a, (a ∪ b)) = 1
4	3	wcomcom 414	. . . 4 C ((a ∪ b), a) = 1
5	4	wcomcom2 415	. . 3 C ((a ∪ b), a⊥ ) = 1
6		wcomorr 412	. . . . . 6 C (b, (b ∪ a)) = 1
7		ax-a2 31	. . . . . . 7 (b ∪ a) = (a ∪ b)
8	7	bi1 118	. . . . . 6 ((b ∪ a) ≡ (a ∪ b)) = 1
9	6, 8	wcbtr 411	. . . . 5 C (b, (a ∪ b)) = 1
10	9	wcomcom 414	. . . 4 C ((a ∪ b), b) = 1
11	10	wcomcom2 415	. . 3 C ((a ∪ b), b⊥ ) = 1
12	5, 11	wfh1 423	. 2 (((a ∪ b) ∩ (a⊥ ∪ b⊥ )) ≡ (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))) = 1
13	2, 12	wr2 371	1 ((a ≡ b)⊥ ≡ (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 8

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-wom 361

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-le 129  df-le1 130  df-le2 131  df-cmtr 134

This theorem is used by: (None)