Theorem eqoreldif 4651

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 487	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ ¬ 𝐴 = 𝐵) → 𝐴 ∈ 𝐶)
2		elsni 4606	. . . . . . 7 ⊢ (𝐴 ∈ {𝐵} → 𝐴 = 𝐵)
3	2	con3i 155	. . . . . 6 ⊢ (¬ 𝐴 = 𝐵 → ¬ 𝐴 ∈ {𝐵})
4	3	adantl 486	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ ¬ 𝐴 = 𝐵) → ¬ 𝐴 ∈ {𝐵})
5	1, 4	eldifd 3916	. . . 4 ⊢ ((𝐴 ∈ 𝐶 ∧ ¬ 𝐴 = 𝐵) → 𝐴 ∈ (𝐶 ∖ {𝐵}))
6	5	ex 417	. . 3 ⊢ (𝐴 ∈ 𝐶 → (¬ 𝐴 = 𝐵 → 𝐴 ∈ (𝐶 ∖ {𝐵})))
7	6	orrd 876	. 2 ⊢ (𝐴 ∈ 𝐶 → (𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵})))
8		eleq1a 2858	. . 3 ⊢ (𝐵 ∈ 𝐶 → (𝐴 = 𝐵 → 𝐴 ∈ 𝐶))
9		eldifi 4085	. . . 4 ⊢ (𝐴 ∈ (𝐶 ∖ {𝐵}) → 𝐴 ∈ 𝐶)
10	9	a1i 11	. . 3 ⊢ (𝐵 ∈ 𝐶 → (𝐴 ∈ (𝐶 ∖ {𝐵}) → 𝐴 ∈ 𝐶))
11	8, 10	jaod 872	. 2 ⊢ (𝐵 ∈ 𝐶 → ((𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵})) → 𝐴 ∈ 𝐶))
12	7, 11	impbid2 229	1 ⊢ (𝐵 ∈ 𝐶 → (𝐴 ∈ 𝐶 ↔ (𝐴 = 𝐵 ∨ 𝐴 ∈ (𝐶 ∖ {𝐵}))))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   ∨ wo 860   = wceq 1570   ∈ wcel 2143   ∖ cdif 3902  {csn 4589

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3457  df-dif 3908  df-sn 4590

This theorem is referenced by:  lcmfunsnlem2  16693