Theorem ud1 595

Description: Unified disjunction for Sasaki implication.

Assertion

Ref	Expression
ud1	(a ∪ b) = ((a →1 b) →1 (((a →1 b) →1 (b →1 a)) →1 a))

Proof of Theorem ud1

Step	Hyp	Ref	Expression
1		ud1lem1 560	. . . . . 6 ((a →1 b) →1 (b →1 a)) = (a ∪ (a⊥ ∩ b⊥ ))
2	1	ud1lem0b 256	. . . . 5 (((a →1 b) →1 (b →1 a)) →1 a) = ((a ∪ (a⊥ ∩ b⊥ )) →1 a)
3		ud1lem2 561	. . . . 5 ((a ∪ (a⊥ ∩ b⊥ )) →1 a) = (a ∪ b)
4	2, 3	ax-r2 36	. . . 4 (((a →1 b) →1 (b →1 a)) →1 a) = (a ∪ b)
5	4	ud1lem0a 255	. . 3 ((a →1 b) →1 (((a →1 b) →1 (b →1 a)) →1 a)) = ((a →1 b) →1 (a ∪ b))
6		ud1lem3 562	. . 3 ((a →1 b) →1 (a ∪ b)) = (a ∪ b)
7	5, 6	ax-r2 36	. 2 ((a →1 b) →1 (((a →1 b) →1 (b →1 a)) →1 a)) = (a ∪ b)
8	7	ax-r1 35	1 (a ∪ b) = ((a →1 b) →1 (((a →1 b) →1 (b →1 a)) →1 a))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from 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-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

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