Theorem reutru 48682

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 1543	. . . 4 ⊢ ⊤
2	1	biantru 529	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ⊤))
3	2	eubii 2583	. 2 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
4		df-reu 3364	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ⊤ ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
5	3, 4	bitr4i 278	1 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Syntax hints:   ↔ wb 206   ∧ wa 395  ⊤wtru 1540   ∈ wcel 2107  ∃!weu 2566  ∃!wreu 3361

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-mo 2538  df-eu 2567  df-reu 3364

This theorem is referenced by: (None)