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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 4737 | . . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴) | |
2 | 1 | eqcomd 2827 | . . 3 ⊢ (𝐵 ∈ 𝐴 → 𝐴 = ((𝐴 ∖ {𝐵}) ∪ {𝐵})) |
3 | 2 | raleqdv 3416 | . 2 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑)) |
4 | ralunb 4167 | . . 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 4607 | . . 3 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ {𝐵}𝜑 ↔ 𝜓)) |
8 | 7 | anbi2d 630 | . 2 ⊢ (𝐵 ∈ 𝐴 → ((∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑) ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓))) |
9 | 3, 5, 8 | 3bitrd 307 | 1 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 208 ∧ wa 398 = wceq 1533 ∈ wcel 2110 ∀wral 3138 ∖ cdif 3933 ∪ cun 3934 {csn 4561 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1907 ax-6 1966 ax-7 2011 ax-8 2112 ax-9 2120 ax-10 2141 ax-11 2156 ax-12 2172 ax-ext 2793 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3an 1085 df-tru 1536 df-ex 1777 df-nf 1781 df-sb 2066 df-clab 2800 df-cleq 2814 df-clel 2893 df-nfc 2963 df-ral 3143 df-rab 3147 df-v 3497 df-sbc 3773 df-dif 3939 df-un 3941 df-in 3943 df-ss 3952 df-nul 4292 df-sn 4562 |
This theorem is referenced by: bj-0int 34387 |
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