Theorem elabd 3665

Description: Explicit demonstration the class {𝑥 ∣ 𝜓} is not empty by the example 𝐴. (Contributed by RP, 12-Aug-2020.) (Revised by AV, 23-Mar-2024.)

Hypotheses

Ref	Expression
elabd.1	⊢ (𝜑 → 𝐴 ∈ 𝑉)
elabd.2	⊢ (𝜑 → 𝜒)
elabd.3	⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
elabd	⊢ (𝜑 → 𝐴 ∈ {𝑥 ∣ 𝜓})

Distinct variable groups:   𝑥,𝐴   𝜒,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)   𝑉(𝑥)

Proof of Theorem elabd

Step	Hyp	Ref	Expression
1		elabd.2	. 2 ⊢ (𝜑 → 𝜒)
2		elabd.1	. . 3 ⊢ (𝜑 → 𝐴 ∈ 𝑉)
3		elabd.3	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜒))
4	3	elabg 3660	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))
5	2, 4	syl 17	. 2 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))
6	1, 5	mpbird 257	1 ⊢ (𝜑 → 𝐴 ∈ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1540   ∈ wcel 2109  {cab 2714

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810

This theorem is referenced by:  intidg  5437  wfrlem15OLD  8342  lubval  18371  glbval  18384  sursubmefmnd  18879  injsubmefmnd  18880  nosupfv  27675  branmfn  32091  orvcval  34495  r1peuqusdeg1  35670  sticksstones3  42166  rngunsnply  43160  hoidmvlelem1  46591  cfsetsnfsetf  47054  iinfconstbaslem  48999