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Theorem oa3-3lem 981
 Description: Lemma for 3-OA(3). Equivalence with substitution into 6-OA dual.
Assertion
Ref Expression
oa3-3lem (a ∩ (a ∪ (b ∩ (((ab) ∪ (ab )) ∪ (((a ∩ 1) ∪ (ac)) ∩ ((b ∩ 1) ∪ (bc))))))) = (a ∩ (a ∪ (b ∩ ((ab) ∪ ((a1 c) ∩ (b1 c))))))

Proof of Theorem oa3-3lem
StepHypRef Expression
1 dfb 94 . . . . . 6 (ab) = ((ab) ∪ (ab ))
21ax-r1 35 . . . . 5 ((ab) ∪ (ab )) = (ab)
3 an1 106 . . . . . . . . 9 (a ∩ 1) = a
4 ax-a1 30 . . . . . . . . 9 a = a
53, 4ax-r2 36 . . . . . . . 8 (a ∩ 1) = a
65ax-r5 38 . . . . . . 7 ((a ∩ 1) ∪ (ac)) = (a ∪ (ac))
7 df-i1 44 . . . . . . . 8 (a1 c) = (a ∪ (ac))
87ax-r1 35 . . . . . . 7 (a ∪ (ac)) = (a1 c)
96, 8ax-r2 36 . . . . . 6 ((a ∩ 1) ∪ (ac)) = (a1 c)
10 an1 106 . . . . . . . . 9 (b ∩ 1) = b
11 ax-a1 30 . . . . . . . . 9 b = b
1210, 11ax-r2 36 . . . . . . . 8 (b ∩ 1) = b
1312ax-r5 38 . . . . . . 7 ((b ∩ 1) ∪ (bc)) = (b ∪ (bc))
14 df-i1 44 . . . . . . . 8 (b1 c) = (b ∪ (bc))
1514ax-r1 35 . . . . . . 7 (b ∪ (bc)) = (b1 c)
1613, 15ax-r2 36 . . . . . 6 ((b ∩ 1) ∪ (bc)) = (b1 c)
179, 162an 79 . . . . 5 (((a ∩ 1) ∪ (ac)) ∩ ((b ∩ 1) ∪ (bc))) = ((a1 c) ∩ (b1 c))
182, 172or 72 . . . 4 (((ab) ∪ (ab )) ∪ (((a ∩ 1) ∪ (ac)) ∩ ((b ∩ 1) ∪ (bc)))) = ((ab) ∪ ((a1 c) ∩ (b1 c)))
1918lan 77 . . 3 (b ∩ (((ab) ∪ (ab )) ∪ (((a ∩ 1) ∪ (ac)) ∩ ((b ∩ 1) ∪ (bc))))) = (b ∩ ((ab) ∪ ((a1 c) ∩ (b1 c))))
2019lor 70 . 2 (a ∪ (b ∩ (((ab) ∪ (ab )) ∪ (((a ∩ 1) ∪ (ac)) ∩ ((b ∩ 1) ∪ (bc)))))) = (a ∪ (b ∩ ((ab) ∪ ((a1 c) ∩ (b1 c)))))
2120lan 77 1 (a ∩ (a ∪ (b ∩ (((ab) ∪ (ab )) ∪ (((a ∩ 1) ∪ (ac)) ∩ ((b ∩ 1) ∪ (bc))))))) = (a ∩ (a ∪ (b ∩ ((ab) ∪ ((a1 c) ∩ (b1 c))))))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 8   →1 wi1 12 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 This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44 This theorem is referenced by:  oa3-6to3  987
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