Theorem bj-raldifsn 37459

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 4748	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)
2	1	eqcomd 2746	. . 3 ⊢ (𝐵 ∈ 𝐴 → 𝐴 = ((𝐴 ∖ {𝐵}) ∪ {𝐵}))
3	2	raleqdv 3298	. 2 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ ((𝐴 ∖ {𝐵}) ∪ {𝐵})𝜑))
4		ralunb 4133	. . 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 4614	. . 3 ⊢ (𝐵 ∈ 𝐴 → (∀𝑥 ∈ {𝐵}𝜑 ↔ 𝜓))
8	7	anbi2d 636	. 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)))

Syntax hints:   → wi 4   ↔ wb 207   ∧ wa 396   = wceq 1547   ∈ wcel 2119  ∀wral 3054   ∖ cdif 3887   ∪ cun 3888  {csn 4562

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-ral 3055  df-rex 3065  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4269  df-sn 4563

This theorem is referenced by:  bj-0int  37460