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| Mirrors > Home > QLE Home > Th. List > ud3lem0c | GIF version | ||
| Description: Lemma for unified disjunction. (Contributed by NM, 22-Nov-1997.) |
| Ref | Expression |
|---|---|
| ud3lem0c | (a →3 b)⊥ = (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ni31 250 | 1 (a →3 b)⊥ = (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) |
| Colors of variables: term |
| Syntax hints: = wb 1 ⊥ wn 4 ∪ wo 6 ∩ wa 7 →3 wi3 14 |
| This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 |
| This theorem depends on definitions: df-a 40 df-i3 46 |
| This theorem is referenced by: ud3lem1a 566 ud3lem1b 567 ud3lem1c 568 ud3lem3a 572 ud3lem3b 573 ud3lem3c 574 ud3lem3 576 u3lem14mp 794 |
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