Theorem rexss 4058

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.)

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem rexss

Step	Hyp	Ref	Expression
1		df-ss 3967	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		pm3.41 492	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝑥 ∈ 𝐵))
3	2	pm4.71rd 562	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
4	3	alexbii 1832	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
5	1, 4	sylbi 217	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6		df-rex 3070	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
7		df-rex 3070	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑)))
8	5, 6, 7	3bitr4g 314	1 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1537  ∃wex 1778   ∈ wcel 2107  ∃wrex 3069   ⊆ wss 3950

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-rex 3070  df-ss 3967

This theorem is referenced by:  oddnn02np1  16386  oddge22np1  16387  evennn02n  16388  evennn2n  16389  2lgslem1a  27436  omssubadd  34303  rexabso  44991  limsupmnfuzlem  45746  sbgoldbo  47779