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Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-raldifsn | Structured version Visualization version GIF version |
Description: All elements in a set satisfy a given property if and only if all but one satisfy that property and that one also does. Typically, this can be used for characterizations that are proved using different methods for a given element and for all others, for instance zero and nonzero numbers, or the empty set and nonempty sets. (Contributed by BJ, 7-Dec-2021.) |
Ref | Expression |
---|---|
bj-raldifsn.is | ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜓)) |
Ref | Expression |
---|---|
bj-raldifsn | ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | difsnid 4735 | . . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴) | |
2 | 1 | eqcomd 2824 | . . 3 ⊢ (𝐵 ∈ 𝐴 → 𝐴 = ((𝐴 ∖ {𝐵}) ∪ {𝐵})) |
3 | 2 | raleqdv 3413 | . 2 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑)) |
4 | ralunb 4164 | . . 3 ⊢ (∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑)) | |
5 | 4 | a1i 11 | . 2 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑))) |
6 | bj-raldifsn.is | . . . 4 ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜓)) | |
7 | 6 | ralsng 4605 | . . 3 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ {𝐵}𝜑 ↔ 𝜓)) |
8 | 7 | anbi2d 628 | . 2 ⊢ (𝐵 ∈ 𝐴 → ((∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑) ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓))) |
9 | 3, 5, 8 | 3bitrd 306 | 1 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 207 ∧ wa 396 = wceq 1528 ∈ wcel 2105 ∀wral 3135 ∖ cdif 3930 ∪ cun 3931 {csn 4557 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1787 ax-4 1801 ax-5 1902 ax-6 1961 ax-7 2006 ax-8 2107 ax-9 2115 ax-10 2136 ax-11 2151 ax-12 2167 ax-ext 2790 |
This theorem depends on definitions: df-bi 208 df-an 397 df-or 842 df-3an 1081 df-tru 1531 df-ex 1772 df-nf 1776 df-sb 2061 df-clab 2797 df-cleq 2811 df-clel 2890 df-nfc 2960 df-ral 3140 df-rab 3144 df-v 3494 df-sbc 3770 df-dif 3936 df-un 3938 df-in 3940 df-ss 3949 df-nul 4289 df-sn 4558 |
This theorem is referenced by: bj-0int 34287 |
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