Theorem sac 835

Description: Theorem showing "Sasaki complement" is an operation. (Contributed by NM, 3-Jan-1999.)

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 256	. 2 ((a →1 c) →1 c) = ((b →1 c) →1 c)
3		u1lem12 781	. 2 ((a →1 c) →1 c) = (a⊥ →1 c)
4		u1lem12 781	. 2 ((b →1 c) →1 c) = (b⊥ →1 c)
5	2, 3, 4	3tr2 64	1 (a⊥ →1 c) = (b⊥ →1 c)

Syntax hints:   = wb 1  ^⊥ wn 4   →₁ wi1 12

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by:  negantlem9  859  negant3  860  negant4  862