Theorem ud1lem0b 248

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 36	. . 3 a⊥ = b⊥
3	1	ran 71	. . 3 (a ∩ c) = (b ∩ c)
4	2, 3	2or 67	. 2 (a⊥ ∪ (a ∩ c)) = (b⊥ ∪ (b ∩ c))
5		df-i1 43	. 2 (a →1 c) = (a⊥ ∪ (a ∩ c))
6		df-i1 43	. 2 (b →1 c) = (b⊥ ∪ (b ∩ c))
7	4, 5, 6	3tr1 60	1 (a →1 c) = (b →1 c)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem is referenced by:  ud1lem0ab 249  wql1 285  ud1 577  oi3oa3lem1 714  oi3oa3 715  u1lem12 763  1oaiii 805  sac 817  oa4to4u 953  oa4uto4g 955  oa4gto4u 956

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-i1 43

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