Theorem ss2rab 3343

Description: Restricted abstraction classes in a subclass relationship. (Contributed by NM, 30-May-1999.)

Assertion

Ref	Expression
ss2rab	⊢ ({x ∈ A ∣ φ} ⊆ {x ∈ A ∣ ψ} ↔ ∀x ∈ A (φ → ψ))

Proof of Theorem ss2rab

Step	Hyp	Ref	Expression
1		df-rab 2624	. . 3 ⊢ {x ∈ A ∣ φ} = {x ∣ (x ∈ A ∧ φ)}
2		df-rab 2624	. . 3 ⊢ {x ∈ A ∣ ψ} = {x ∣ (x ∈ A ∧ ψ)}
3	1, 2	sseq12i 3298	. 2 ⊢ ({x ∈ A ∣ φ} ⊆ {x ∈ A ∣ ψ} ↔ {x ∣ (x ∈ A ∧ φ)} ⊆ {x ∣ (x ∈ A ∧ ψ)})
4		ss2ab 3335	. 2 ⊢ ({x ∣ (x ∈ A ∧ φ)} ⊆ {x ∣ (x ∈ A ∧ ψ)} ↔ ∀x((x ∈ A ∧ φ) → (x ∈ A ∧ ψ)))
5		df-ral 2620	. . 3 ⊢ (∀x ∈ A (φ → ψ) ↔ ∀x(x ∈ A → (φ → ψ)))
6		imdistan 670	. . . 4 ⊢ ((x ∈ A → (φ → ψ)) ↔ ((x ∈ A ∧ φ) → (x ∈ A ∧ ψ)))
7	6	albii 1566	. . 3 ⊢ (∀x(x ∈ A → (φ → ψ)) ↔ ∀x((x ∈ A ∧ φ) → (x ∈ A ∧ ψ)))
8	5, 7	bitr2i 241	. 2 ⊢ (∀x((x ∈ A ∧ φ) → (x ∈ A ∧ ψ)) ↔ ∀x ∈ A (φ → ψ))
9	3, 4, 8	3bitri 262	1 ⊢ ({x ∈ A ∣ φ} ⊆ {x ∈ A ∣ ψ} ↔ ∀x ∈ A (φ → ψ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   ∈ wcel 1710  {cab 2339  ∀wral 2615  {crab 2619   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ral 2620  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is used by:  ss2rabdv  3348  ss2rabi  3349