Metamath Proof Explorer


Theorem rexss

Description: Restricted existential quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015) Avoid axioms. (Revised by SN, 14-Oct-2025)

Ref Expression
Assertion rexss ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 df-ss ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
2 pm3.41 ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) → 𝑥 ∈ 𝐵 ) )
3 2 pm4.71rd ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ( 𝑥 ∈ 𝐵 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
4 3 alexbii ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
5 1 4 sylbi ⊢ ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
6 df-rex ⊢ ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) )
7 df-rex ⊢ ( ∃ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐵 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
8 5 6 7 3bitr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐵 ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )