Theorem wle0 372

Description: 0 is l.e. anything.

Assertion

Ref	Expression
wle0	(0 ≤2 a) = 1

Proof of Theorem wle0

Step	Hyp	Ref	Expression
1		df-le 121	. 2 (0 ≤2 a) = ((0 ∪ a) ≡ a)
2		ax-a2 30	. . . 4 (0 ∪ a) = (a ∪ 0)
3		or0 94	. . . 4 (a ∪ 0) = a
4	2, 3	ax-r2 35	. . 3 (0 ∪ a) = a
5	4	bi1 110	. 2 ((0 ∪ a) ≡ a) = 1
6	1, 5	ax-r2 35	1 (0 ≤2 a) = 1

Syntax hints:   = wb 1   ≡ tb 5   ∪ wo 6  1wt 9  0wf 10   ≤₂ wle2 11

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-le 121

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