Theorem rextru 46149

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

Assertion

Ref	Expression
rextru	⊢ (∃𝑥 𝑥 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 ⊤)

Proof of Theorem rextru

Step	Hyp	Ref	Expression
1		tru 1543	. . . 4 ⊢ ⊤
2	1	biantru 530	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ⊤))
3	2	exbii 1850	. 2 ⊢ (∃𝑥 𝑥 ∈ 𝐴 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
4		df-rex 3070	. 2 ⊢ (∃𝑥 ∈ 𝐴 ⊤ ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
5	3, 4	bitr4i 277	1 ⊢ (∃𝑥 𝑥 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 ⊤)

Syntax hints:   ↔ wb 205   ∧ wa 396  ⊤wtru 1540  ∃wex 1782   ∈ wcel 2106  ∃wrex 3065

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-rex 3070

This theorem is referenced by:  reutruALT  46152