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Theorem euequ 2684
Description: There exists a unique set equal to a given set. Special case of eueqi 3685 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 3684 and eueqi 3685 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.
StepHypRef Expression
1 ax6ev 1973 . 2 𝑥 𝑥 = 𝑦
2 ax6ev 1973 . . 3 𝑧 𝑧 = 𝑦
3 equeuclr 2031 . . . 4 (𝑧 = 𝑦 → (𝑥 = 𝑦𝑥 = 𝑧))
43alrimiv 1929 . . 3 (𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑦𝑥 = 𝑧))
52, 4eximii 1838 . 2 𝑧𝑥(𝑥 = 𝑦𝑥 = 𝑧)
6 eu3v 2656 . 2 (∃!𝑥 𝑥 = 𝑦 ↔ (∃𝑥 𝑥 = 𝑦 ∧ ∃𝑧𝑥(𝑥 = 𝑦𝑥 = 𝑧)))
71, 5, 6mpbir2an 710 1 ∃!𝑥 𝑥 = 𝑦
Colors of variables: wff setvar class
Syntax hints:  wi 4  wal 1536  wex 1781  ∃!weu 2654
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016
This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-mo 2624  df-eu 2655
This theorem is referenced by:  axsepgfromrep  5182  copsexgw  5362  copsexg  5363  oprabidw  7169  oprabid  7170
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