Theorem elscottab 9947

Description: An element of the output of the Scott operation applied to a class abstraction satisfies the class abstraction's predicate. (Contributed by Rohan Ridenour, 14-Aug-2023.)

Hypothesis

Ref	Expression
elscottab.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elscottab	⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝜓)

Distinct variable groups:   𝜓,𝑥   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑦)

Proof of Theorem elscottab

Step	Hyp	Ref	Expression
1		scottss 9940	. . 3 ⊢ Scott {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜑}
2	1	sseli 3905	. 2 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝑦 ∈ {𝑥 ∣ 𝜑})
3		vex 3454	. . 3 ⊢ 𝑦 ∈ V
4		elscottab.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	3, 4	elab 3632	. 2 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6	2, 5	sylib 221	1 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∈ wcel 2144  {cab 2738  Scott cscott 9933

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 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3897  df-scott 9934

This theorem is used by:  karden  9964  cpcolld  45349  grucollcld  45351