Theorem eu6im 2605

Description: One direction of eu6 2604 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 2035	. . 3 ⊢ (∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥𝜑)
2	1	anim1i 627	. 2 ⊢ ((∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)) → (∃𝑥𝜑 ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
3		eu6lem 2603	. 2 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ (∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
4		eu3v 2600	. 2 ⊢ (∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
5	2, 3, 4	3imtr4i 295	1 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) → ∃!𝑥𝜑)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812  ∃!weu 2598

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-mo 2569  df-eu 2599

This theorem is used by:  tz6.12-2  6872  eufsnlem  49678  termcarweu  50365