Theorem elscottab 9866

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 9865	. . 3 ⊢ Scott {𝑥 ∣ 𝜑} ⊆ {𝑥 ∣ 𝜑}
2	1	sseli 3941	. 2 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝑦 ∈ {𝑥 ∣ 𝜑})
3		vex 3467	. . 3 ⊢ 𝑦 ∈ V
4		elscottab.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	3, 4	elab 3647	. 2 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6	2, 5	sylib 221	1 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝜑} → 𝜓)

Syntax hints:   → wi 4   ↔ wb 209   ∈ wcel 2149  {cab 2747  Scott cscott 9853

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  df-rab 3424  df-v 3465  df-ss 3930  df-scott 9854

This theorem is referenced by:  cpcolld  44855  grucollcld  44857