Theorem eu6im 2569

Description: One direction of eu6 2568 needs fewer axioms. (Contributed by Wolf Lammen, 2-Mar-2023.)

Assertion

Ref	Expression
eu6im	⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) → ∃!𝑥𝜑)

Distinct variable groups:   𝑥,𝑦   𝜑,𝑦

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem eu6im

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		exsbim 2005	. . 3 ⊢ (∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥𝜑)
2	1	anim1i 615	. 2 ⊢ ((∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)) → (∃𝑥𝜑 ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
3		eu6lem 2567	. 2 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ (∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
4		eu3v 2564	. 2 ⊢ (∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
5	2, 3, 4	3imtr4i 291	1 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) → ∃!𝑥𝜑)

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1539  ∃wex 1781  ∃!weu 2562

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 1913  ax-6 1971  ax-7 2011

This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1782  df-mo 2534  df-eu 2563

This theorem is referenced by:  eufsnlem  47595