Theorem eujust 2206

Description: A soundness justification theorem for df-eu 2208, showing that the definition is equivalent to itself with its dummy variable renamed. Note that y and z needn't be distinct variables. See eujustALT 2207 for a proof that provides an example of how it can be achieved through the use of dvelim 2016. (Contributed by NM, 11-Mar-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)

Assertion

Ref	Expression
eujust	⊢ (∃y∀x(φ ↔ x = y) ↔ ∃z∀x(φ ↔ x = z))

Distinct variable groups:   x,y   x,z   φ,y   φ,z

Allowed substitution hint:   φ(x)

Proof of Theorem eujust

Dummy variable w is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ2 1686	. . . . 5 ⊢ (y = w → (x = y ↔ x = w))
2	1	bibi2d 309	. . . 4 ⊢ (y = w → ((φ ↔ x = y) ↔ (φ ↔ x = w)))
3	2	albidv 1625	. . 3 ⊢ (y = w → (∀x(φ ↔ x = y) ↔ ∀x(φ ↔ x = w)))
4	3	cbvexv 2003	. 2 ⊢ (∃y∀x(φ ↔ x = y) ↔ ∃w∀x(φ ↔ x = w))
5		equequ2 1686	. . . . 5 ⊢ (w = z → (x = w ↔ x = z))
6	5	bibi2d 309	. . . 4 ⊢ (w = z → ((φ ↔ x = w) ↔ (φ ↔ x = z)))
7	6	albidv 1625	. . 3 ⊢ (w = z → (∀x(φ ↔ x = w) ↔ ∀x(φ ↔ x = z)))
8	7	cbvexv 2003	. 2 ⊢ (∃w∀x(φ ↔ x = w) ↔ ∃z∀x(φ ↔ x = z))
9	4, 8	bitri 240	1 ⊢ (∃y∀x(φ ↔ x = y) ↔ ∃z∀x(φ ↔ x = z))

Syntax hints:   ↔ wb 176  ∀wal 1540  ∃wex 1541   = wceq 1642

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is referenced by: (None)