Theorem eqoreldif 4689

Description: An element of a set is either equal to another element of the set or a member of the difference of the set and the singleton containing the other element. (Contributed by AV, 25-Aug-2020.) (Proof shortened by JJ, 23-Jul-2021.)

Assertion

Ref	Expression
eqoreldif	⊢ (𝐵 ∈ 𝐶 → (𝐴 ∈ 𝐶 ↔ (𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵}))))

Proof of Theorem eqoreldif

Step	Hyp	Ref	Expression
1		simpl 482	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ ¬ 𝐴 = 𝐵) → 𝐴 ∈ 𝐶)
2		elsni 4646	. . . . . . 7 ⊢ (𝐴 ∈ {𝐵} → 𝐴 = 𝐵)
3	2	con3i 154	. . . . . 6 ⊢ (¬ 𝐴 = 𝐵 → ¬ 𝐴 ∈ {𝐵})
4	3	adantl 481	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ ¬ 𝐴 = 𝐵) → ¬ 𝐴 ∈ {𝐵})
5	1, 4	eldifd 3960	. . . 4 ⊢ ((𝐴 ∈ 𝐶 ∧ ¬ 𝐴 = 𝐵) → 𝐴 ∈ (𝐶 ∖ {𝐵}))
6	5	ex 412	. . 3 ⊢ (𝐴 ∈ 𝐶 → (¬ 𝐴 = 𝐵 → 𝐴 ∈ (𝐶 ∖ {𝐵})))
7	6	orrd 860	. 2 ⊢ (𝐴 ∈ 𝐶 → (𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵})))
8		eleq1a 2827	. . 3 ⊢ (𝐵 ∈ 𝐶 → (𝐴 = 𝐵 → 𝐴 ∈ 𝐶))
9		eldifi 4127	. . . 4 ⊢ (𝐴 ∈ (𝐶 ∖ {𝐵}) → 𝐴 ∈ 𝐶)
10	9	a1i 11	. . 3 ⊢ (𝐵 ∈ 𝐶 → (𝐴 ∈ (𝐶 ∖ {𝐵}) → 𝐴 ∈ 𝐶))
11	8, 10	jaod 856	. 2 ⊢ (𝐵 ∈ 𝐶 → ((𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵})) → 𝐴 ∈ 𝐶))
12	7, 11	impbid2 225	1 ⊢ (𝐵 ∈ 𝐶 → (𝐴 ∈ 𝐶 ↔ (𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵}))))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 844   = wceq 1540   ∈ wcel 2105   ∖ cdif 3946  {csn 4629

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 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3475  df-dif 3952  df-sn 4630

This theorem is referenced by:  lcmfunsnlem2  16582