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| Mirrors > Home > MPE Home > Th. List > difpr | Structured version Visualization version 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 4128 | . . 3 ⊢ {𝐵, 𝐶} = ({𝐵} ∪ {𝐶}) | |
| 2 | 1 | difeq2i 3687 | . 2 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = (𝐴 ∖ ({𝐵} ∪ {𝐶})) |
| 3 | difun1 3846 | . 2 ⊢ (𝐴 ∖ ({𝐵} ∪ {𝐶})) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) | |
| 4 | 2, 3 | eqtri 2632 | 1 ⊢ (𝐴 ∖ {𝐵, 𝐶}) = ((𝐴 ∖ {𝐵}) ∖ {𝐶}) |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1475 ∖ cdif 3537 ∪ cun 3538 {csn 4125 {cpr 4127 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
| This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ral 2901 df-rab 2905 df-v 3175 df-dif 3543 df-un 3545 df-in 3547 df-pr 4128 |
| This theorem is referenced by: hashdifpr 13064 nbgrassvwo2 25967 nbgrssvwo2 40587 nbupgrres 40592 nbupgruvtxres 40634 uvtxupgrres 40635 |
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