Theorem 3oa3 1025

Description: 3-variable orthoarguesion law expressed with the 3OA identity abbreviation. (Contributed by NM, 27-May-2004.)

Assertion

Ref	Expression
3oa3	((a →1 c) ∩ (a ≡ c ≡OA b)) ≤ (b →1 c)

Proof of Theorem 3oa3

Step	Hyp	Ref	Expression
1		df-id3oa 57	. . 3 (a ≡ c ≡OA b) = (((a →1 c) ∩ (b →1 c)) ∪ ((a⊥ →1 c) ∩ (b⊥ →1 c)))
2	1	lan 77	. 2 ((a →1 c) ∩ (a ≡ c ≡OA b)) = ((a →1 c) ∩ (((a →1 c) ∩ (b →1 c)) ∪ ((a⊥ →1 c) ∩ (b⊥ →1 c))))
3		3oa2 1024	. 2 ((a →1 c) ∩ (((a →1 c) ∩ (b →1 c)) ∪ ((a⊥ →1 c) ∩ (b⊥ →1 c)))) ≤ (b →1 c)
4	2, 3	bltr 138	1 ((a →1 c) ∩ (a ≡ c ≡OA b)) ≤ (b →1 c)

This proof depends on syntax axioms:   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   ≡ wid3oa 27

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  ax-3oa 998

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-id3oa 57  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)