Theorem reutru 47124

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 1545	. . . 4 ⊢ ⊤
2	1	biantru 530	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ⊤))
3	2	eubii 2578	. 2 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
4		df-reu 3376	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ⊤ ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
5	3, 4	bitr4i 277	1 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Syntax hints:   ↔ wb 205   ∧ wa 396  ⊤wtru 1542   ∈ wcel 2106  ∃!weu 2561  ∃!wreu 3373

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-mo 2533  df-eu 2562  df-reu 3376

This theorem is referenced by: (None)