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Theorem gomaex3lem3 916
 Description: Lemma for Godowski 6-var -> Mayet Example 3.
Assertion
Ref Expression
gomaex3lem3 ((p1 q) ∪ (pq)) = p

Proof of Theorem gomaex3lem3
StepHypRef Expression
1 anor1 88 . . . . 5 (p ∩ (pq) ) = (p ∪ (pq))
21ax-r1 35 . . . 4 (p ∪ (pq)) = (p ∩ (pq) )
3 df-i1 44 . . . . 5 (p1 q) = (p ∪ (pq))
43ax-r4 37 . . . 4 (p1 q) = (p ∪ (pq))
5 id 59 . . . 4 (p ∩ (pq) ) = (p ∩ (pq) )
62, 4, 53tr1 63 . . 3 (p1 q) = (p ∩ (pq) )
76ax-r5 38 . 2 ((p1 q) ∪ (pq)) = ((p ∩ (pq) ) ∪ (pq))
8 coman1 185 . . 3 (pq) C p
9 comid 187 . . . 4 (pq) C (pq)
109comcom2 183 . . 3 (pq) C (pq)
118, 10fh3r 475 . 2 ((p ∩ (pq) ) ∪ (pq)) = ((p ∪ (pq)) ∩ ((pq) ∪ (pq)))
12 orabs 120 . . . 4 (p ∪ (pq)) = p
13 ax-a2 31 . . . . 5 ((pq) ∪ (pq)) = ((pq) ∪ (pq) )
14 df-t 41 . . . . . 6 1 = ((pq) ∪ (pq) )
1514ax-r1 35 . . . . 5 ((pq) ∪ (pq) ) = 1
1613, 15ax-r2 36 . . . 4 ((pq) ∪ (pq)) = 1
1712, 162an 79 . . 3 ((p ∪ (pq)) ∩ ((pq) ∪ (pq))) = (p ∩ 1)
18 an1 106 . . 3 (p ∩ 1) = p
1917, 18ax-r2 36 . 2 ((p ∪ (pq)) ∩ ((pq) ∪ (pq))) = p
207, 11, 193tr 65 1 ((p1 q) ∪ (pq)) = p
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →1 wi1 12 This theorem was proved from 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 This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133 This theorem is referenced by:  gomaex3lem7  920
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