Theorem oal2 999

Description: Orthoarguesian law - →₂ version. (Contributed by NM, 20-Jul-1999.)

Assertion

Ref	Expression
oal2	((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) ≤ (a →2 c)

Proof of Theorem oal2

Step	Hyp	Ref	Expression
1		ax-3oa 998	. 2 ((b⊥ →1 a⊥ ) ∩ ((b⊥ ∩ c⊥ ) ∪ ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ )))) ≤ (c⊥ →1 a⊥ )
2		i2i1 267	. . 3 (a →2 b) = (b⊥ →1 a⊥ )
3		anor3 90	. . . . 5 (b⊥ ∩ c⊥ ) = (b ∪ c)⊥
4	3	ax-r1 35	. . . 4 (b ∪ c)⊥ = (b⊥ ∩ c⊥ )
5		i2i1 267	. . . . 5 (a →2 c) = (c⊥ →1 a⊥ )
6	2, 5	2an 79	. . . 4 ((a →2 b) ∩ (a →2 c)) = ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ ))
7	4, 6	2or 72	. . 3 ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c))) = ((b⊥ ∩ c⊥ ) ∪ ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ )))
8	2, 7	2an 79	. 2 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) = ((b⊥ →1 a⊥ ) ∩ ((b⊥ ∩ c⊥ ) ∪ ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ ))))
9	1, 8, 5	le3tr1 140	1 ((a →2 b) ∩ ((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a →2 c)))) ≤ (a →2 c)

This proof depends on syntax axioms:   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   →₂ wi2 13

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-3oa 998

This proof depends on definitions:  df-a 40  df-i1 44  df-i2 45  df-le1 130  df-le2 131

This theorem is used by:  oal1  1000  oaliii  1001  oagen2  1016  mloa  1018  oadistc0  1021