Theorem mojust 2539

Description: Soundness justification theorem for df-mo 2540 (note that 𝑦 and 𝑧 need not be disjoint, although the weaker theorem with that disjoint variable condition added would be enough to justify the soundness of the definition). (Contributed by NM, 11-Mar-2010.) Added this theorem by adapting the proof of eujust 2571. (Revised by BJ, 30-Sep-2022.)

Assertion

Ref	Expression
mojust	⊢ (∃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 mojust

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ2 2025	. . . . 5 ⊢ (𝑦 = 𝑡 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑡))
2	1	imbi2d 340	. . . 4 ⊢ (𝑦 = 𝑡 → ((𝜑 → 𝑥 = 𝑦) ↔ (𝜑 → 𝑥 = 𝑡)))
3	2	albidv 1920	. . 3 ⊢ (𝑦 = 𝑡 → (∀𝑥(𝜑 → 𝑥 = 𝑦) ↔ ∀𝑥(𝜑 → 𝑥 = 𝑡)))
4	3	cbvexvw 2036	. 2 ⊢ (∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦) ↔ ∃𝑡∀𝑥(𝜑 → 𝑥 = 𝑡))
5		equequ2 2025	. . . . 5 ⊢ (𝑡 = 𝑧 → (𝑥 = 𝑡 ↔ 𝑥 = 𝑧))
6	5	imbi2d 340	. . . 4 ⊢ (𝑡 = 𝑧 → ((𝜑 → 𝑥 = 𝑡) ↔ (𝜑 → 𝑥 = 𝑧)))
7	6	albidv 1920	. . 3 ⊢ (𝑡 = 𝑧 → (∀𝑥(𝜑 → 𝑥 = 𝑡) ↔ ∀𝑥(𝜑 → 𝑥 = 𝑧)))
8	7	cbvexvw 2036	. 2 ⊢ (∃𝑡∀𝑥(𝜑 → 𝑥 = 𝑡) ↔ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧))
9	4, 8	bitri 275	1 ⊢ (∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦) ↔ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538  ∃wex 1779

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780

This theorem is referenced by: (None)