Theorem eujust 2460

Description: A soundness justification theorem for df-eu 2462, 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 2461 for a proof that provides an example of how it can be achieved through the use of dvelim 2325. (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 1940	. . . . 5 ⊢ (𝑦 = 𝑤 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑤))
2	1	bibi2d 331	. . . 4 ⊢ (𝑦 = 𝑤 → ((𝜑 ↔ 𝑥 = 𝑦) ↔ (𝜑 ↔ 𝑥 = 𝑤)))
3	2	albidv 1836	. . 3 ⊢ (𝑦 = 𝑤 → (∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑤)))
4	3	cbvexv 2263	. 2 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∃𝑤∀𝑥(𝜑 ↔ 𝑥 = 𝑤))
5		equequ2 1940	. . . . 5 ⊢ (𝑤 = 𝑧 → (𝑥 = 𝑤 ↔ 𝑥 = 𝑧))
6	5	bibi2d 331	. . . 4 ⊢ (𝑤 = 𝑧 → ((𝜑 ↔ 𝑥 = 𝑤) ↔ (𝜑 ↔ 𝑥 = 𝑧)))
7	6	albidv 1836	. . 3 ⊢ (𝑤 = 𝑧 → (∀𝑥(𝜑 ↔ 𝑥 = 𝑤) ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑧)))
8	7	cbvexv 2263	. 2 ⊢ (∃𝑤∀𝑥(𝜑 ↔ 𝑥 = 𝑤) ↔ ∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧))
9	4, 8	bitri 263	1 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧))

Syntax hints:   ↔ wb 195  ∀wal 1473  ∃wex 1695

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-nf 1701

This theorem is referenced by: (None)