Theorem ud1lem0b 256

Description: Introduce →₁ to the right.

Hypothesis

Ref	Expression
ud1lem0a.1	a = b

Assertion

Ref	Expression
ud1lem0b	(a →1 c) = (b →1 c)

Proof of Theorem ud1lem0b

Step	Hyp	Ref	Expression
1		ud1lem0a.1	. . . 4 a = b
2	1	ax-r4 37	. . 3 a⊥ = b⊥
3	1	ran 78	. . 3 (a ∩ c) = (b ∩ c)
4	2, 3	2or 72	. 2 (a⊥ ∪ (a ∩ c)) = (b⊥ ∪ (b ∩ c))
5		df-i1 44	. 2 (a →1 c) = (a⊥ ∪ (a ∩ c))
6		df-i1 44	. 2 (b →1 c) = (b⊥ ∪ (b ∩ c))
7	4, 5, 6	3tr1 63	1 (a →1 c) = (b →1 c)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from axioms:  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-i1 44

This theorem is referenced by:  ud1lem0ab  257  wql1  293  ud1  595  oi3oa3lem1  732  oi3oa3  733  u1lem12  781  1oaiii  823  sac  835  oa4to4u  973  oa4uto4g  975  oa4gto4u  976