Theorem bj-raldifsn 37113

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 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
bj-raldifsn	⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓)))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem bj-raldifsn

Step	Hyp	Ref	Expression
1		difsnid 4760	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
2	1	eqcomd 2736	. . 3 ⊢ (𝐵 ∈ 𝐴 → 𝐴 = ((𝐴 ∖ {𝐵}) ∪ {𝐵}))
3	2	raleqdv 3290	. 2 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑))
4		ralunb 4145	. . 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 4626	. . 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 305	1 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ (∀𝑥 ∈ (𝐴 ∖ {𝐵})𝜑 ∧ 𝜓)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1541   ∈ wcel 2110  ∀wral 3045   ∖ cdif 3897   ∪ cun 3898  {csn 4574

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2112  ax-9 2120  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2722  df-clel 2804  df-ral 3046  df-rex 3055  df-rab 3394  df-v 3436  df-dif 3903  df-un 3905  df-in 3907  df-ss 3917  df-nul 4282  df-sn 4575

This theorem is referenced by:  bj-0int  37114