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| Mirrors > Home > QLE Home > Th. List > oasr | GIF version | ||
| Description: Reverse of oas 925 lemma for studying the orthoarguesian law. (Contributed by NM, 28-Dec-1998.) |
| Ref | Expression |
|---|---|
| oasr.1 | ((a →1 c) ∩ (a ∪ b)) ≤ c |
| Ref | Expression |
|---|---|
| oasr | (a⊥ ∩ (a ∪ b)) ≤ c |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | u1lem9b 778 | . . 3 a⊥ ≤ (a →1 c) | |
| 2 | 1 | leran 153 | . 2 (a⊥ ∩ (a ∪ b)) ≤ ((a →1 c) ∩ (a ∪ b)) |
| 3 | oasr.1 | . 2 ((a →1 c) ∩ (a ∪ b)) ≤ c | |
| 4 | 2, 3 | letr 137 | 1 (a⊥ ∩ (a ∪ b)) ≤ c |
| 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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