Theorem bj-raldifsn 37344

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 4768	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
2	1	eqcomd 2743	. . 3 ⊢ (𝐵 ∈ 𝐴 → 𝐴 = ((𝐴 ∖ {𝐵}) ∪ {𝐵}))
3	2	raleqdv 3298	. 2 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑))
4		ralunb 4151	. . 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 4634	. . 3 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ {𝐵}𝜑 ↔ 𝜓))
8	7	anbi2d 631	. 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 1542   ∈ wcel 2114  ∀wral 3052   ∖ cdif 3900   ∪ cun 3901  {csn 4582

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-sn 4583

This theorem is referenced by:  bj-0int  37345