Theorem negant0 839

Description: Negated antecedent identity.

Hypothesis

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

Assertion

Ref	Expression
negant0	(a⊥ →0 c) = (b⊥ →0 c)

Proof of Theorem negant0

Step	Hyp	Ref	Expression
1		negant.1	. . . 4 (a →1 c) = (b →1 c)
2	1	negantlem7 837	. . 3 (a ∪ c) = (b ∪ c)
3		ax-a1 29	. . . 4 a = a⊥ ⊥
4	3	ax-r5 37	. . 3 (a ∪ c) = (a⊥ ⊥ ∪ c)
5		ax-a1 29	. . . 4 b = b⊥ ⊥
6	5	ax-r5 37	. . 3 (b ∪ c) = (b⊥ ⊥ ∪ c)
7	2, 4, 6	3tr2 61	. 2 (a⊥ ⊥ ∪ c) = (b⊥ ⊥ ∪ c)
8		df-i0 42	. 2 (a⊥ →0 c) = (a⊥ ⊥ ∪ c)
9		df-i0 42	. 2 (b⊥ →0 c) = (b⊥ ⊥ ∪ c)
10	7, 8, 9	3tr1 60	1 (a⊥ →0 c) = (b⊥ →0 c)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   →₀ wi0 12   →₁ wi1 13

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-i0 42  df-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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