Theorem womle 290

Description: An equality implying the WOM law.

Hypothesis

Ref	Expression
womle.1	(a ∩ (a →1 b)) = (a ∩ (a →2 b))

Assertion

Ref	Expression
womle	((a →2 b)⊥ ∪ (a →1 b)) = 1

Proof of Theorem womle

Step	Hyp	Ref	Expression
1		womle.1	. . . . 5 (a ∩ (a →1 b)) = (a ∩ (a →2 b))
2	1	ax-r1 34	. . . 4 (a ∩ (a →2 b)) = (a ∩ (a →1 b))
3		lear 153	. . . 4 (a ∩ (a →1 b)) ≤ (a →1 b)
4	2, 3	bltr 130	. . 3 (a ∩ (a →2 b)) ≤ (a →1 b)
5		leor 151	. . 3 (a →1 b) ≤ ((a →2 b)⊥ ∪ (a →1 b))
6	4, 5	letr 129	. 2 (a ∩ (a →2 b)) ≤ ((a →2 b)⊥ ∪ (a →1 b))
7	6	womle2a 287	1 ((a →2 b)⊥ ∪ (a →1 b)) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₁ wi1 13   →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

This theorem depends on definitions:  df-a 39  df-t 40  df-f 41  df-i1 43  df-le1 122  df-le2 123

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