Theorem ud4lem0a 262

Description: Introduce →₄ to the left. (Contributed by NM, 23-Nov-1997.)

Hypothesis

Ref	Expression
ud4lem0a.1	a = b

Assertion

Ref	Expression
ud4lem0a	(c →4 a) = (c →4 b)

Proof of Theorem ud4lem0a

Step	Hyp	Ref	Expression
1		ud4lem0a.1	. . . . 5 a = b
2	1	lan 77	. . . 4 (c ∩ a) = (c ∩ b)
3	1	lan 77	. . . 4 (c⊥ ∩ a) = (c⊥ ∩ b)
4	2, 3	2or 72	. . 3 ((c ∩ a) ∪ (c⊥ ∩ a)) = ((c ∩ b) ∪ (c⊥ ∩ b))
5	1	lor 70	. . . 4 (c⊥ ∪ a) = (c⊥ ∪ b)
6	1	ax-r4 37	. . . 4 a⊥ = b⊥
7	5, 6	2an 79	. . 3 ((c⊥ ∪ a) ∩ a⊥ ) = ((c⊥ ∪ b) ∩ b⊥ )
8	4, 7	2or 72	. 2 (((c ∩ a) ∪ (c⊥ ∩ a)) ∪ ((c⊥ ∪ a) ∩ a⊥ )) = (((c ∩ b) ∪ (c⊥ ∩ b)) ∪ ((c⊥ ∪ b) ∩ b⊥ ))
9		df-i4 47	. 2 (c →4 a) = (((c ∩ a) ∪ (c⊥ ∩ a)) ∪ ((c⊥ ∪ a) ∩ a⊥ ))
10		df-i4 47	. 2 (c →4 b) = (((c ∩ b) ∪ (c⊥ ∩ b)) ∪ ((c⊥ ∪ b) ∩ b⊥ ))
11	8, 9, 10	3tr1 63	1 (c →4 a) = (c →4 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 15

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

This proof depends on definitions:  df-a 40  df-i4 47

This theorem is used by:  i4i3  271  nom43  328  ud4  598