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Theorem bj-raldifsn 37101
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.)
Hypothesis
Ref Expression
bj-raldifsn.is (𝑥 = 𝐵 → (𝜑𝜓))
Assertion
Ref Expression
bj-raldifsn (𝐵𝐴 → (∀𝑥𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑𝜓)))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem bj-raldifsn
StepHypRef Expression
1 difsnid 4810 . . . 4 (𝐵𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
21eqcomd 2743 . . 3 (𝐵𝐴𝐴 = ((𝐴 ∖ {𝐵}) ∪ {𝐵}))
32raleqdv 3326 . 2 (𝐵𝐴 → (∀𝑥𝐴 𝜑 ↔ ∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑))
4 ralunb 4197 . . 3 (∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑))
54a1i 11 . 2 (𝐵𝐴 → (∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑)))
6 bj-raldifsn.is . . . 4 (𝑥 = 𝐵 → (𝜑𝜓))
76ralsng 4675 . . 3 (𝐵𝐴 → (∀𝑥 ∈ {𝐵}𝜑𝜓))
87anbi2d 630 . 2 (𝐵𝐴 → ((∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ ∀𝑥 ∈ {𝐵}𝜑) ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑𝜓)))
93, 5, 83bitrd 305 1 (𝐵𝐴 → (∀𝑥𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑𝜓)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1540  wcel 2108  wral 3061  cdif 3948  cun 3949  {csn 4626
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-sn 4627
This theorem is referenced by:  bj-0int  37102
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