Theorem reutru 48782

Description: Two ways of expressing "exactly one" element. (Contributed by Zhi Wang, 23-Sep-2024.)

Assertion

Ref	Expression
reutru	⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Proof of Theorem reutru

Step	Hyp	Ref	Expression
1		tru 1544	. . . 4 ⊢ ⊤
2	1	biantru 529	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ⊤))
3	2	eubii 2579	. 2 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
4		df-reu 3357	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ⊤ ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
5	3, 4	bitr4i 278	1 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Syntax hints:   ↔ wb 206   ∧ wa 395  ⊤wtru 1541   ∈ wcel 2109  ∃!weu 2562  ∃!wreu 3354

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-mo 2534  df-eu 2563  df-reu 3357

This theorem is referenced by:  isinito2lem  49467