Theorem sac 817

Description: Theorem showing "Sasaki complement" is an operation.

Hypothesis

Ref	Expression
sac.1	(a →1 c) = (b →1 c)

Assertion

Ref	Expression
sac	(a⊥ →1 c) = (b⊥ →1 c)

Proof of Theorem sac

Step	Hyp	Ref	Expression
1		sac.1	. . 3 (a →1 c) = (b →1 c)
2	1	ud1lem0b 248	. 2 ((a →1 c) →1 c) = ((b →1 c) →1 c)
3		u1lem12 763	. 2 ((a →1 c) →1 c) = (a⊥ →1 c)
4		u1lem12 763	. 2 ((b →1 c) →1 c) = (b⊥ →1 c)
5	2, 3, 4	3tr2 61	1 (a⊥ →1 c) = (b⊥ →1 c)

Syntax hints:   = wb 1  ^⊥ wn 4   →₁ wi1 13

This theorem is referenced by:  negantlem9 841  negant3 842  negant4 844

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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