Theorem ralssOLD 3983

Description: Obsolete version of ralss 3981 as of 14-Oct-2025. (Contributed by Stefan O'Rear, 3-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
ralssOLD	⊢ (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 ralssOLD

Step	Hyp	Ref	Expression
1		ssel 3902	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	pm4.71rd 572	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
3	2	imbi1d 344	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 → 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑)))
4		impexp 456	. . 3 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
5	3, 4	bitrdi 290	. 2 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
6	5	ralbidv2 3181	1 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2144  ∀wral 3076   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-clel 2835  df-ral 3077  df-ss 3896

This theorem is used by: (None)