Theorem euequ 2625

Description: There exists a unique set equal to a given set. Special case of eueqi 3670 proved using only predicate calculus. The proof needs 𝑦 = 𝑧 be free of 𝑥. This is ensured by having 𝑥 and 𝑦 be distinct. Alternately, a distinctor ¬ ∀𝑥𝑥 = 𝑦 could have been used instead. See eueq 3669 and eueqi 3670 for classes. (Contributed by Stefan Allan, 4-Dec-2008.) (Proof shortened by Wolf Lammen, 8-Sep-2019.) Reduce axiom usage. (Revised by Wolf Lammen, 1-Mar-2023.)

Assertion

Ref	Expression
euequ	⊢ ∃!𝑥 𝑥 = 𝑦

Distinct variable group:   𝑥,𝑦

Proof of Theorem euequ

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ax6ev 2004	. 2 ⊢ ∃𝑥 𝑥 = 𝑦
2		ax6ev 2004	. . 3 ⊢ ∃𝑧 𝑧 = 𝑦
3		equeuclr 2058	. . . 4 ⊢ (𝑧 = 𝑦 → (𝑥 = 𝑦 → 𝑥 = 𝑧))
4	3	alrimiv 1962	. . 3 ⊢ (𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧))
5	2, 4	eximii 1870	. 2 ⊢ ∃𝑧∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧)
6		eu3v 2598	. 2 ⊢ (∃!𝑥 𝑥 = 𝑦 ↔ (∃𝑥 𝑥 = 𝑦 ∧ ∃𝑧∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧)))
7	1, 5, 6	mpbir2an 724	1 ⊢ ∃!𝑥 𝑥 = 𝑦

This proof depends on syntax axioms:   → wi 4  ∀wal 1568  ∃wex 1812  ∃!weu 2596

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-mo 2567  df-eu 2597

This theorem is used by:  axsepgfromrep  5250  copsexgwOLD  5465  oprabidw  7471  oprabid  7472