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Theorem orordir 113
Description: Distribution of disjunction over disjunction.
Assertion
Ref Expression
orordir ((ab) ∪ c) = ((ac) ∪ (bc))

Proof of Theorem orordir
StepHypRef Expression
1 oridm 110 . . . 4 (cc) = c
21ax-r1 35 . . 3 c = (cc)
32lor 70 . 2 ((ab) ∪ c) = ((ab) ∪ (cc))
4 or4 84 . 2 ((ab) ∪ (cc)) = ((ac) ∪ (bc))
53, 4ax-r2 36 1 ((ab) ∪ c) = ((ac) ∪ (bc))
Colors of variables: term
Syntax hints:   = wb 1  wo 6
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-t 41  df-f 42
This theorem is referenced by:  leror  152  wql2lem2  289  wleror  393  ska2  432
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