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Theorem euequ1 2624
Description: Equality has existential uniqueness. Special case of eueq1 3531 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.
StepHypRef Expression
1 ax6evr 2100 . . 3 𝑧 𝑦 = 𝑧
2 equequ2 2111 . . . 4 (𝑦 = 𝑧 → (𝑥 = 𝑦𝑥 = 𝑧))
32alrimiv 2007 . . 3 (𝑦 = 𝑧 → ∀𝑥(𝑥 = 𝑦𝑥 = 𝑧))
41, 3eximii 1912 . 2 𝑧𝑥(𝑥 = 𝑦𝑥 = 𝑧)
5 df-eu 2622 . 2 (∃!𝑥 𝑥 = 𝑦 ↔ ∃𝑧𝑥(𝑥 = 𝑦𝑥 = 𝑧))
64, 5mpbir 221 1 ∃!𝑥 𝑥 = 𝑦
Colors of variables: wff setvar class
Syntax hints:  wb 196  wal 1629  wex 1852  ∃!weu 2618
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093
This theorem depends on definitions:  df-bi 197  df-an 383  df-ex 1853  df-eu 2622
This theorem is referenced by:  copsexg  5083  oprabid  6822
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