Theorem euequ1 2613

Description: Equality has existential uniqueness. Special case of eueq1 3520 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. (Contributed by Stefan Allan, 4-Dec-2008.) (Proof shortened by Wolf Lammen, 8-Sep-2019.)

Assertion

Ref	Expression
euequ1	⊢ ∃!𝑥 𝑥 = 𝑦

Distinct variable group:   𝑥,𝑦

Proof of Theorem euequ1

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ax6evr 2097	. . 3 ⊢ ∃𝑧 𝑦 = 𝑧
2		equequ2 2108	. . . 4 ⊢ (𝑦 = 𝑧 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑧))
3	2	alrimiv 2004	. . 3 ⊢ (𝑦 = 𝑧 → ∀𝑥(𝑥 = 𝑦 ↔ 𝑥 = 𝑧))
4	1, 3	eximii 1913	. 2 ⊢ ∃𝑧∀𝑥(𝑥 = 𝑦 ↔ 𝑥 = 𝑧)
5		df-eu 2611	. 2 ⊢ (∃!𝑥 𝑥 = 𝑦 ↔ ∃𝑧∀𝑥(𝑥 = 𝑦 ↔ 𝑥 = 𝑧))
6	4, 5	mpbir 221	1 ⊢ ∃!𝑥 𝑥 = 𝑦

Syntax hints:   ↔ wb 196  ∀wal 1630  ∃wex 1853  ∃!weu 2607

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090

This theorem depends on definitions:  df-bi 197  df-an 385  df-ex 1854  df-eu 2611

This theorem is referenced by:  copsexg  5104  oprabid  6840