Theorem alsconv 44258

Description: There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.)

Assertion

Ref	Expression
alsconv	⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑)

Proof of Theorem alsconv

Step	Hyp	Ref	Expression
1		df-ral 3087	. . 3 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
2	1	anbi1i 614	. 2 ⊢ ((∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴))
3		df-alsc 44257	. 2 ⊢ (∀!𝑥 ∈ 𝐴𝜑 ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴))
4		df-alsi 44256	. 2 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴))
5	2, 3, 4	3bitr4ri 296	1 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑)

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 387  ∀wal 1505  ∃wex 1742   ∈ wcel 2050  ∀wral 3082  ∀!walsi 44254  ∀!walsc 44255

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

This theorem depends on definitions:  df-bi 199  df-an 388  df-ral 3087  df-alsi 44256  df-alsc 44257

This theorem is referenced by: (None)