Theorem elabd 3680

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 3675	. . 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 1539   ∈ wcel 2107  {cab 2713

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815

This theorem is referenced by:  intidg  5461  wfrlem15OLD  8364  lubval  18402  glbval  18415  sursubmefmnd  18910  injsubmefmnd  18911  nosupfv  27752  branmfn  32125  orvcval  34461  r1peuqusdeg1  35649  sticksstones3  42150  rngunsnply  43186  hoidmvlelem1  46615  cfsetsnfsetf  47075