Metamath Proof Explorer


Theorem bj-equs45fv

Description: Version of equs45f with a disjoint variable condition, which does not require ax-13 . Note that the version of equs5 with a disjoint variable condition is actually sbalex (up to adding a superfluous antecedent). (Contributed by BJ, 11-Sep-2019) (Proof modification is discouraged.)

Ref Expression
Hypothesis bj-equs45fv.1 ⊢ Ⅎ 𝑦 𝜑
Assertion bj-equs45fv ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )

Proof

Step Hyp Ref Expression
1 bj-equs45fv.1 ⊢ Ⅎ 𝑦 𝜑
2 1 nf5ri ⊢ ( 𝜑 → ∀ 𝑦 𝜑 )
3 2 anim2i ⊢ ( ( 𝑥 = 𝑦 ∧ 𝜑 ) → ( 𝑥 = 𝑦 ∧ ∀ 𝑦 𝜑 ) )
4 3 eximi ⊢ ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ ∀ 𝑦 𝜑 ) )
5 equs5av ⊢ ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ ∀ 𝑦 𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )
6 4 5 syl ⊢ ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )
7 equs4v ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) )
8 6 7 impbii ⊢ ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )