Theorem intmin4 4937

Description: Elimination of a conjunct in a class intersection. (Contributed by NM, 31-Jul-2006.)

Assertion

Ref	Expression
intmin4	⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} = ∩ {𝑥 ∣ 𝜑})

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem intmin4

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssintab 4925	. . . 4 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝜑 → 𝐴 ⊆ 𝑥))
2		simpr 490	. . . . . . . 8 ⊢ ((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝜑)
3		ancr 556	. . . . . . . 8 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → (𝜑 → (𝐴 ⊆ 𝑥 ∧ 𝜑)))
4	2, 3	impbid2 229	. . . . . . 7 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → ((𝐴 ⊆ 𝑥 ∧ 𝜑) ↔ 𝜑))
5	4	imbi1d 344	. . . . . 6 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → (((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)))
6	5	alimi 1844	. . . . 5 ⊢ (∀𝑥(𝜑 → 𝐴 ⊆ 𝑥) → ∀𝑥(((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)))
7		albi 1851	. . . . 5 ⊢ (∀𝑥(((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)) → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
8	6, 7	syl 18	. . . 4 ⊢ (∀𝑥(𝜑 → 𝐴 ⊆ 𝑥) → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
9	1, 8	sylbi 220	. . 3 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
10		vex 3455	. . . 4 ⊢ 𝑦 ∈ V
11	10	elintab 4919	. . 3 ⊢ (𝑦 ∈ ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} ↔ ∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥))
12	10	elintab 4919	. . 3 ⊢ (𝑦 ∈ ∩ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥))
13	9, 11, 12	3bitr4g 317	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → (𝑦 ∈ ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} ↔ 𝑦 ∈ ∩ {𝑥 ∣ 𝜑}))
14	13	eqrdv 2759	1 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} = ∩ {𝑥 ∣ 𝜑})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570   ∈ wcel 2145  {cab 2739   ⊆ wss 3899  ∩ cint 4907

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-v 3453  df-ss 3916  df-int 4908

This theorem is used by: (None)