Theorem ralss 4010

Description: Restricted universal 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
ralss	⊢ (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 ralss

Step	Hyp	Ref	Expression
1		df-ss 3922	. . . 4 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		id 23	. . . . . . . 8 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	2	pm4.71rd 571	. . . . . . 7 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
4	3	imbi1d 344	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 → 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑)))
5		impexp 455	. . . . . 6 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
6	4, 5	bitrdi 290	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
7	6	alimi 1841	. . . 4 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
8	1, 7	sylbi 220	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
9		albi 1848	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))) → (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
10	8, 9	syl 18	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
11		df-ral 3080	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
12		df-ral 3080	. 2 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
13	10, 11, 12	3bitr4g 317	1 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1568   ∈ wcel 2143  ∀wral 3079   ⊆ wss 3905

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839

This theorem depends on definitions:  df-bi 210  df-an 401  df-ral 3080  df-ss 3922

This theorem is referenced by:  acsfn  17710  acsfn1  17712  acsfn2  17714  acsfn1p  20902  mdetunilem9  22777  ntrneik3  44842  ntrneix3  44843  ntrneik13  44844  ntrneix13  44845  ralabso  45697