Theorem nelrdva 3666

Description: Deduce negative membership from an implication. (Contributed by Thierry Arnoux, 27-Nov-2017.)

Hypothesis

Ref	Expression
nelrdva.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵)

Assertion

Ref	Expression
nelrdva	⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

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

Proof of Theorem nelrdva

Step	Hyp	Ref	Expression
1		eqidd 2764	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 = 𝐵)
2		eleq1 2852	. . . . . . 7 ⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴))
3	2	anbi2d 642	. . . . . 6 ⊢ (𝑥 = 𝐵 → ((𝜑 ∧ 𝑥 ∈ 𝐴) ↔ (𝜑 ∧ 𝐵 ∈ 𝐴)))
4		neeq1 3021	. . . . . 6 ⊢ (𝑥 = 𝐵 → (𝑥 ≠ 𝐵 ↔ 𝐵 ≠ 𝐵))
5	3, 4	imbi12d 347	. . . . 5 ⊢ (𝑥 = 𝐵 → (((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵) ↔ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)))
6		nelrdva.1	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵)
7	5, 6	vtoclg 3521	. . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵))
8	7	anabsi7 684	. . 3 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)
9	8	neneqd 2964	. 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → ¬ 𝐵 = 𝐵)
10	1, 9	pm2.65da 829	1 ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1569   ∈ wcel 2146   ≠ wne 2959

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960

This theorem is used by:  ustfilxp  24547  metustfbas  24891  drngmxidl  34057  dflringlem2  34083  fourierdlem72  47330