Theorem eqab2 38501

Description: Implication of a class abstraction. (Contributed by Peter Mazsa, 16-Apr-2019.)

Assertion

Ref	Expression
eqab2	⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → ∀𝑥 ∈ 𝐴 𝜑)

Proof of Theorem eqab2

Step	Hyp	Ref	Expression
1		biimp 215	. . 3 ⊢ ((𝑥 ∈ 𝐴 ↔ 𝜑) → (𝑥 ∈ 𝐴 → 𝜑))
2	1	alimi 1813	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
3	2	ralrid 3060	1 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → ∀𝑥 ∈ 𝐴 𝜑)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1540   ∈ wcel 2114  ∀wral 3052

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 207  df-ral 3053

This theorem is referenced by:  dmqsblocks  39218