Theorem ud1lem0a 255

Description: Introduce →₁ to the left.

Hypothesis

Ref	Expression
ud1lem0a.1	a = b

Assertion

Ref	Expression
ud1lem0a	(c →1 a) = (c →1 b)

Proof of Theorem ud1lem0a

Step	Hyp	Ref	Expression
1		ud1lem0a.1	. . . 4 a = b
2	1	lan 77	. . 3 (c ∩ a) = (c ∩ b)
3	2	lor 70	. 2 (c⊥ ∪ (c ∩ a)) = (c⊥ ∪ (c ∩ b))
4		df-i1 44	. 2 (c →1 a) = (c⊥ ∪ (c ∩ a))
5		df-i1 44	. 2 (c →1 b) = (c⊥ ∪ (c ∩ b))
6	3, 4, 5	3tr1 63	1 (c →1 a) = (c →1 b)

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  nom42  327  ud1  595  u3lem13b  790  2oai1u  822  1oaiii  823  oa3to4lem1  945  oa3to4lem2  946  oa4to6lem1  960  oa4to6lem2  961  oa4to6lem3  962