Metamath Proof Explorer


Theorem rexabsod

Description: Deduction form of rexabso . (Contributed by Eric Schmidt, 19-Oct-2025)

Ref Expression
Hypothesis ralabsod.1 ⊢ φ → Tr ⁡ M
Assertion rexabsod ⊢ φ ∧ A ∈ M → ∃ x ∈ A ψ ↔ ∃ x ∈ M x ∈ A ∧ ψ

Proof

Step Hyp Ref Expression
1 ralabsod.1 ⊢ φ → Tr ⁡ M
2 rexabso ⊢ Tr ⁡ M ∧ A ∈ M → ∃ x ∈ A ψ ↔ ∃ x ∈ M x ∈ A ∧ ψ
3 1 2 sylan ⊢ φ ∧ A ∈ M → ∃ x ∈ A ψ ↔ ∃ x ∈ M x ∈ A ∧ ψ