Theorem eu6im 2600

Description: One direction of eu6 2599 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 2034	. . 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 2598	. 2 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ (∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑) ∧ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧)))
4		eu3v 2595	. 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 1567  ∃wex 1811  ∃!weu 2593

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-mo 2564  df-eu 2594

This theorem is used by:  tz6.12-2  6877  eufsnlem  50088  termcarweu  50799