Metamath Proof Explorer


Theorem exellimddv

Description: Eliminate an antecedent when the antecedent is elementhood, deduction version. See exellim for the closed form, which requires the use of a universal quantifier. (Contributed by ML, 17-Jul-2020)

Ref Expression
Hypotheses exellimddv.1 ⊢ ( 𝜑 → ∃ 𝑥 𝑥 ∈ 𝐴 )
exellimddv.2 ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → 𝜓 ) )
Assertion exellimddv ( 𝜑 → 𝜓 )

Proof

Step Hyp Ref Expression
1 exellimddv.1 ⊢ ( 𝜑 → ∃ 𝑥 𝑥 ∈ 𝐴 )
2 exellimddv.2 ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → 𝜓 ) )
3 2 alrimiv ⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜓 ) )
4 exellim ⊢ ( ( ∃ 𝑥 𝑥 ∈ 𝐴 ∧ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜓 ) ) → 𝜓 )
5 1 3 4 syl2anc ⊢ ( 𝜑 → 𝜓 )