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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 ) ) )