Theorem euequ1 2269

Description: Equality has existential uniqueness. Special case of eueq1 3241 proved using only predicate calculus. The proof needs y = z be free of x. This is ensured by having x and y be distinct. Alternatevly, a distinctor -. A. x x = 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	|- E! 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		ax6ev 1796	. . 3 |- E. z z = y
2		equequ2 1848	. . . . 5 |- ( y = z -> ( x = y <-> x = z ) )
3	2	equcoms 1844	. . . 4 |- ( z = y -> ( x = y <-> x = z ) )
4	3	alrimiv 1763	. . 3 |- ( z = y -> A. x ( x = y <-> x = z ) )
5	1, 4	eximii 1704	. 2 |- E. z A. x ( x = y <-> x = z )
6		df-eu 2267	. 2 |- ( E! x x = y <-> E. z A. x ( x = y <-> x = z ) )
7	5, 6	mpbir 212	1 |- E! x x = y

Syntax hints:   <-> wb 187   A.wal 1435   E.wex 1659   E!weu 2263

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1665  ax-4 1678  ax-5 1748  ax-6 1794  ax-7 1838

This theorem depends on definitions:  df-bi 188  df-ex 1660  df-eu 2267

This theorem is referenced by:  copsexg  4699  oprabid  6324