Theorem ud5lem0b 257

Description: Introduce →₅ to the right.

Hypothesis

Ref	Expression
ud5lem0a.1	a = b

Assertion

Ref	Expression
ud5lem0b	(a →5 c) = (b →5 c)

Proof of Theorem ud5lem0b

Step	Hyp	Ref	Expression
1		ud5lem0a.1	. . . . 5 a = b
2	1	ran 71	. . . 4 (a ∩ c) = (b ∩ c)
3	1	ax-r4 36	. . . . 5 a⊥ = b⊥
4	3	ran 71	. . . 4 (a⊥ ∩ c) = (b⊥ ∩ c)
5	2, 4	2or 67	. . 3 ((a ∩ c) ∪ (a⊥ ∩ c)) = ((b ∩ c) ∪ (b⊥ ∩ c))
6	3	ran 71	. . 3 (a⊥ ∩ c⊥ ) = (b⊥ ∩ c⊥ )
7	5, 6	2or 67	. 2 (((a ∩ c) ∪ (a⊥ ∩ c)) ∪ (a⊥ ∩ c⊥ )) = (((b ∩ c) ∪ (b⊥ ∩ c)) ∪ (b⊥ ∩ c⊥ ))
8		df-i5 47	. 2 (a →5 c) = (((a ∩ c) ∪ (a⊥ ∩ c)) ∪ (a⊥ ∩ c⊥ ))
9		df-i5 47	. 2 (b →5 c) = (((b ∩ c) ∪ (b⊥ ∩ c)) ∪ (b⊥ ∩ c⊥ ))
10	7, 8, 9	3tr1 60	1 (a →5 c) = (b →5 c)

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

This theorem is referenced by:  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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