Theorem eqab2 39102

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 218	. . 3 ⊢ ((𝑥 ∈ 𝐴 ↔ 𝜑) → (𝑥 ∈ 𝐴 → 𝜑))
2	1	alimi 1844	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
3	2	ralrid 3084	1 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → ∀𝑥 ∈ 𝐴 𝜑)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   ∈ wcel 2145  ∀wral 3076

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-ral 3077

This theorem is used by:  dmqsblocks  39819