Theorem elabd 3649

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 3644	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))
5	2, 4	syl 18	. 2 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))
6	1, 5	mpbird 260	1 ⊢ (𝜑 → 𝐴 ∈ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 209   = wceq 1567   ∈ wcel 2149  {cab 2747

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844

This theorem is referenced by:  intidg  5439  lubval  18410  glbval  18423  sursubmefmnd  18955  injsubmefmnd  18956  nosupfv  27836  branmfn  32398  orvcval  34793  r1peuqusdeg1  36068  sticksstones3  42839  rngunsnply  43822  hoidmvlelem1  47235  cfsetsnfsetf  47718  iinfconstbaslem  49762