Theorem reutru 45767

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 1547	. . . 4 ⊢ ⊤
2	1	biantru 533	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ⊤))
3	2	eubii 2584	. 2 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
4		df-reu 3058	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ⊤ ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
5	3, 4	bitr4i 281	1 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Syntax hints:   ↔ wb 209   ∧ wa 399  ⊤wtru 1544   ∈ wcel 2112  ∃!weu 2567  ∃!wreu 3053

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-ex 1788  df-mo 2539  df-eu 2568  df-reu 3058

This theorem is referenced by: (None)