Theorem ud2 578

Description: Unified disjunction for Dishkant implication.

Assertion

Ref	Expression
ud2	(a ∪ b) = ((a →2 b) →2 (((a →2 b) →2 (b →2 a)) →2 a))

Proof of Theorem ud2

Step	Hyp	Ref	Expression
1		ud2lem1 545	. . . . . 6 ((a →2 b) →2 (b →2 a)) = (a ∪ (a⊥ ∩ b⊥ ))
2	1	ud2lem0b 251	. . . . 5 (((a →2 b) →2 (b →2 a)) →2 a) = ((a ∪ (a⊥ ∩ b⊥ )) →2 a)
3		ud2lem2 546	. . . . 5 ((a ∪ (a⊥ ∩ b⊥ )) →2 a) = (a ∪ b)
4	2, 3	ax-r2 35	. . . 4 (((a →2 b) →2 (b →2 a)) →2 a) = (a ∪ b)
5	4	ud2lem0a 250	. . 3 ((a →2 b) →2 (((a →2 b) →2 (b →2 a)) →2 a)) = ((a →2 b) →2 (a ∪ b))
6		ud2lem3 547	. . 3 ((a →2 b) →2 (a ∪ b)) = (a ∪ b)
7	5, 6	ax-r2 35	. 2 ((a →2 b) →2 (((a →2 b) →2 (b →2 a)) →2 a)) = (a ∪ b)
8	7	ax-r1 34	1 (a ∪ b) = ((a →2 b) →2 (((a →2 b) →2 (b →2 a)) →2 a))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 14

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-i2 44  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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