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Theorem oa4btoc 966
 Description: Derivation of 4-OA law variant.
Hypothesis
Ref Expression
oa4btoc.1 ((a1 g) ∩ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))))) ≤ g
Assertion
Ref Expression
oa4btoc (a ∩ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))))) ≤ g

Proof of Theorem oa4btoc
StepHypRef Expression
1 leo 158 . . . 4 a ≤ (a ∪ (ag))
2 df-i1 44 . . . . 5 (a1 g) = (a ∪ (ag))
32ax-r1 35 . . . 4 (a ∪ (ag)) = (a1 g)
41, 3lbtr 139 . . 3 a ≤ (a1 g)
5 leid 148 . . . . . 6 (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))) ≤ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g))))
65lelor 166 . . . . 5 (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g))))) ≤ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))
76lelan 167 . . . 4 (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))) ≤ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g))))))
87lelor 166 . . 3 (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g))))))) ≤ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))))
94, 8le2an 169 . 2 (a ∩ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))))) ≤ ((a1 g) ∩ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g))))))))
10 oa4btoc.1 . 2 ((a1 g) ∩ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))))) ≤ g
119, 10letr 137 1 (a ∩ (a ∪ (c ∩ (((ac) ∪ ((a1 g) ∩ (c1 g))) ∪ (((ae) ∪ ((a1 g) ∩ (e1 g))) ∩ ((ce) ∪ ((c1 g) ∩ (e1 g)))))))) ≤ g
 Colors of variables: term Syntax hints:   ≤ wle 2  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →1 wi1 12 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131 This theorem is referenced by: (None)
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