Theorem neg3ant1 866

Description: Lemma for negated antecedent identity. (Contributed by NM, 7-Aug-2001.)

Hypothesis

Ref	Expression
neg3ant.1	(a →3 c) = (b →3 c)

Assertion

Ref	Expression
neg3ant1	(a →1 c) = (b →1 c)

Proof of Theorem neg3ant1

Step	Hyp	Ref	Expression
1		neg3ant.1	. . . . . 6 (a →3 c) = (b →3 c)
2	1	neg3antlem2 865	. . . . 5 a⊥ ≤ (b →1 c)
3	1	neg3antlem1 864	. . . . 5 (a ∩ c) ≤ (b →1 c)
4	2, 3	lel2or 170	. . . 4 (a⊥ ∪ (a ∩ c)) ≤ (b →1 c)
5		df-i1 44	. . . 4 (b →1 c) = (b⊥ ∪ (b ∩ c))
6	4, 5	lbtr 139	. . 3 (a⊥ ∪ (a ∩ c)) ≤ (b⊥ ∪ (b ∩ c))
7	1	ax-r1 35	. . . . . 6 (b →3 c) = (a →3 c)
8	7	neg3antlem2 865	. . . . 5 b⊥ ≤ (a →1 c)
9	7	neg3antlem1 864	. . . . 5 (b ∩ c) ≤ (a →1 c)
10	8, 9	lel2or 170	. . . 4 (b⊥ ∪ (b ∩ c)) ≤ (a →1 c)
11		df-i1 44	. . . 4 (a →1 c) = (a⊥ ∪ (a ∩ c))
12	10, 11	lbtr 139	. . 3 (b⊥ ∪ (b ∩ c)) ≤ (a⊥ ∪ (a ∩ c))
13	6, 12	lebi 145	. 2 (a⊥ ∪ (a ∩ c)) = (b⊥ ∪ (b ∩ c))
14	13, 11, 5	3tr1 63	1 (a →1 c) = (b →1 c)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   →₃ wi3 14

This proof depends on 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 proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i3 46  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)