Theorem alsconv 14109

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 2453	. . 3 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
2	1	anbi1i 455	. 2 ⊢ ((∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴))
3		df-alsc 14108	. 2 ⊢ (∀!𝑥 ∈ 𝐴𝜑 ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑥 𝑥 ∈ 𝐴))
4		df-alsi 14107	. 2 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ∧ ∃𝑥 𝑥 ∈ 𝐴))
5	2, 3, 4	3bitr4ri 212	1 ⊢ (∀!𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀!𝑥 ∈ 𝐴𝜑)

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1346  ∃wex 1485   ∈ wcel 2141  ∀wral 2448  ∀!walsi 14105  ∀!walsc 14106

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-ral 2453  df-alsi 14107  df-alsc 14108

This theorem is referenced by: (None)