Theorem ka4lem 229

Description: Lemma for KA4 soundness (AND version) - uses OL only. (Contributed by NM, 25-Oct-1997.)

Assertion

Ref	Expression
ka4lem	((a ∩ b)⊥ ∪ ((a ∩ c) ≡ (b ∩ c))) = 1

Proof of Theorem ka4lem

Step	Hyp	Ref	Expression
1		df-a 40	. . . 4 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
2	1	con2 67	. . 3 (a ∩ b)⊥ = (a⊥ ∪ b⊥ )
3		df-a 40	. . . . 5 (a ∩ c) = (a⊥ ∪ c⊥ )⊥
4		df-a 40	. . . . 5 (b ∩ c) = (b⊥ ∪ c⊥ )⊥
5	3, 4	2bi 99	. . . 4 ((a ∩ c) ≡ (b ∩ c)) = ((a⊥ ∪ c⊥ )⊥ ≡ (b⊥ ∪ c⊥ )⊥ )
6		conb 122	. . . . 5 ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ )) = ((a⊥ ∪ c⊥ )⊥ ≡ (b⊥ ∪ c⊥ )⊥ )
7	6	ax-r1 35	. . . 4 ((a⊥ ∪ c⊥ )⊥ ≡ (b⊥ ∪ c⊥ )⊥ ) = ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ ))
8	5, 7	ax-r2 36	. . 3 ((a ∩ c) ≡ (b ∩ c)) = ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ ))
9	2, 8	2or 72	. 2 ((a ∩ b)⊥ ∪ ((a ∩ c) ≡ (b ∩ c))) = ((a⊥ ∪ b⊥ ) ∪ ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ )))
10		ka4lemo 228	. 2 ((a⊥ ∪ b⊥ ) ∪ ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ ))) = 1
11	9, 10	ax-r2 36	1 ((a ∩ b)⊥ ∪ ((a ∩ c) ≡ (b ∩ c))) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 8

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

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131

This theorem is used by: (None)