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Theorem oal1 1000
Description: Orthoarguesian law - →1 version derived from →1 version.
Assertion
Ref Expression
oal1 ((a1 c) ∩ ((ab) ∪ ((a1 c) ∩ (b1 c)))) ≤ (b1 c)

Proof of Theorem oal1
StepHypRef Expression
1 oal2 999 . 2 ((c2 a ) ∩ ((ab ) ∪ ((c2 a ) ∩ (c2 b )))) ≤ (c2 b )
2 i1i2 266 . . 3 (a1 c) = (c2 a )
3 df-a 40 . . . 4 (ab) = (ab )
4 i1i2 266 . . . . 5 (b1 c) = (c2 b )
52, 42an 79 . . . 4 ((a1 c) ∩ (b1 c)) = ((c2 a ) ∩ (c2 b ))
63, 52or 72 . . 3 ((ab) ∪ ((a1 c) ∩ (b1 c))) = ((ab ) ∪ ((c2 a ) ∩ (c2 b )))
72, 62an 79 . 2 ((a1 c) ∩ ((ab) ∪ ((a1 c) ∩ (b1 c)))) = ((c2 a ) ∩ ((ab ) ∪ ((c2 a ) ∩ (c2 b ))))
81, 7, 4le3tr1 140 1 ((a1 c) ∩ ((ab) ∪ ((a1 c) ∩ (b1 c)))) ≤ (b1 c)
Colors of variables: term
Syntax hints:   ≤ wle 2   wn 4   ∪ wo 6   ∩ wa 7   →1 wi1 12   →2 wi2 13
This theorem was proved from 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 theorem depends on definitions:  df-a 40  df-i1 44  df-i2 45  df-le1 130  df-le2 131
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