Theorem elscottab 44688

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 44687	. . 3 ⊢ Scott {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜑}
2	1	sseli 3911	. 2 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝑦 ∈ {𝑥 ∣ 𝜑})
3		vex 3435	. . 3 ⊢ 𝑦 ∈ V
4		elscottab.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	3, 4	elab 3617	. 2 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6	2, 5	sylib 219	1 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝜓)

Syntax hints:   → wi 4   ↔ wb 207   ∈ wcel 2119  {cab 2717  Scott cscott 44679

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-rab 3392  df-v 3433  df-ss 3900  df-scott 44680

This theorem is referenced by:  cpcolld  44702  grucollcld  44704