Theorem reutru 49425

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 1564	. . . 4 ⊢ ⊤
2	1	biantru 537	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ⊤))
3	2	eubii 2612	. 2 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
4		df-reu 3368	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ⊤ ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ⊤))
5	3, 4	bitr4i 280	1 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Syntax hints:   ↔ wb 208   ∧ wa 399  ⊤wtru 1561   ∈ wcel 2142  ∃!weu 2595  ∃!wreu 3365

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1563  df-ex 1800  df-mo 2566  df-eu 2596  df-reu 3368

This theorem is referenced by:  isinito2lem  50119