Theorem ud5lem0a 256

Description: Introduce →₅ to the left.

Hypothesis

Ref	Expression
ud5lem0a.1	a = b

Assertion

Ref	Expression
ud5lem0a	(c →5 a) = (c →5 b)

Proof of Theorem ud5lem0a

Step	Hyp	Ref	Expression
1		ud5lem0a.1	. . . . 5 a = b
2	1	lan 70	. . . 4 (c ∩ a) = (c ∩ b)
3	1	lan 70	. . . 4 (c⊥ ∩ a) = (c⊥ ∩ b)
4	2, 3	2or 67	. . 3 ((c ∩ a) ∪ (c⊥ ∩ a)) = ((c ∩ b) ∪ (c⊥ ∩ b))
5	1	ax-r4 36	. . . 4 a⊥ = b⊥
6	5	lan 70	. . 3 (c⊥ ∩ a⊥ ) = (c⊥ ∩ b⊥ )
7	4, 6	2or 67	. 2 (((c ∩ a) ∪ (c⊥ ∩ a)) ∪ (c⊥ ∩ a⊥ )) = (((c ∩ b) ∪ (c⊥ ∩ b)) ∪ (c⊥ ∩ b⊥ ))
8		df-i5 47	. 2 (c →5 a) = (((c ∩ a) ∪ (c⊥ ∩ a)) ∪ (c⊥ ∩ a⊥ ))
9		df-i5 47	. 2 (c →5 b) = (((c ∩ b) ∪ (c⊥ ∩ b)) ∪ (c⊥ ∩ b⊥ ))
10	7, 8, 9	3tr1 60	1 (c →5 a) = (c →5 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 17

This theorem is referenced by:  nom45 322  ud5 581

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

This theorem depends on definitions:  df-a 39  df-i5 47

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