Theorem conb 122

Description: Contraposition law. (Contributed by NM, 10-Aug-1997.)

Assertion

Ref	Expression
conb	(a ≡ b) = (a⊥ ≡ b⊥ )

Proof of Theorem conb

Step	Hyp	Ref	Expression
1		ax-a2 31	. . 3 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))
2		ax-a1 30	. . . . 5 a = a⊥ ⊥
3		ax-a1 30	. . . . 5 b = b⊥ ⊥
4	2, 3	2an 79	. . . 4 (a ∩ b) = (a⊥ ⊥ ∩ b⊥ ⊥ )
5	4	lor 70	. . 3 ((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ⊥ ∩ b⊥ ⊥ ))
6	1, 5	ax-r2 36	. 2 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ⊥ ∩ b⊥ ⊥ ))
7		dfb 94	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
8		dfb 94	. 2 (a⊥ ≡ b⊥ ) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ⊥ ∩ b⊥ ⊥ ))
9	6, 7, 8	3tr1 63	1 (a ≡ b) = (a⊥ ≡ b⊥ )

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

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-b 39  df-a 40

This theorem is used by:  di  126  wr4  199  wcon  202  wcon1  207  wcon2  208  wwfh3  218  wwfh4  219  ka4lem  229  ska3  232  nomcon5  306  nom55  336  wom2  434  u3lemax4  796  comanbn  873