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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 3590 | . . 3 ⊢ {𝐵, 𝐶} = ({𝐵} ∪ {𝐶}) | |
2 | 1 | difeq2i 3242 | . 2 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶})) |
3 | difun1 3387 | . 2 ⊢ (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) | |
4 | 2, 3 | eqtri 2191 | 1 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) |
Colors of variables: wff set class |
Syntax hints: = wceq 1348 ∖ cdif 3118 ∪ cun 3119 {csn 3583 {cpr 3584 |
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 609 ax-in2 610 ax-io 704 ax-5 1440 ax-7 1441 ax-gen 1442 ax-ie1 1486 ax-ie2 1487 ax-8 1497 ax-10 1498 ax-11 1499 ax-i12 1500 ax-bndl 1502 ax-4 1503 ax-17 1519 ax-i9 1523 ax-ial 1527 ax-i5r 1528 ax-ext 2152 |
This theorem depends on definitions: df-bi 116 df-tru 1351 df-nf 1454 df-sb 1756 df-clab 2157 df-cleq 2163 df-clel 2166 df-nfc 2301 df-ral 2453 df-rab 2457 df-v 2732 df-dif 3123 df-un 3125 df-in 3127 df-pr 3590 |
This theorem is referenced by: hashdifpr 10755 |
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