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Mirrors > Home > ILE Home > Th. List > difpr | GIF version |
Description: Removing two elements as pair of elements corresponds to removing each of the two elements as singletons. (Contributed by Alexander van der Vekens, 13-Jul-2018.) |
Ref | Expression |
---|---|
difpr | ⊢ (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-pr 3529 | . . 3 ⊢ {𝐵, 𝐶} = ({𝐵} ∪ {𝐶}) | |
2 | 1 | difeq2i 3186 | . 2 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶})) |
3 | difun1 3331 | . 2 ⊢ (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) | |
4 | 2, 3 | eqtri 2158 | 1 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) |
Colors of variables: wff set class |
Syntax hints: = wceq 1331 ∖ cdif 3063 ∪ cun 3064 {csn 3522 {cpr 3523 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 603 ax-in2 604 ax-io 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-bndl 1486 ax-4 1487 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2119 |
This theorem depends on definitions: df-bi 116 df-tru 1334 df-nf 1437 df-sb 1736 df-clab 2124 df-cleq 2130 df-clel 2133 df-nfc 2268 df-ral 2419 df-rab 2423 df-v 2683 df-dif 3068 df-un 3070 df-in 3072 df-pr 3529 |
This theorem is referenced by: hashdifpr 10559 |
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