Theorem unissel 4900

Description: Condition turning a subclass relationship for union into an equality. (Contributed by NM, 18-Jul-2006.)

Assertion

Ref	Expression
unissel	⊢ ((∪ 𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝐴) → ∪ 𝐴 = 𝐵)

Proof of Theorem unissel

Step	Hyp	Ref	Expression
1		simpl 488	. 2 ⊢ ((∪ 𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝐴) → ∪ 𝐴 ⊆ 𝐵)
2		elssuni 4899	. . 3 ⊢ (𝐵 ∈ 𝐴 → 𝐵 ⊆ ∪ 𝐴)
3	2	adantl 487	. 2 ⊢ ((∪ 𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝐴) → 𝐵 ⊆ ∪ 𝐴)
4	1, 3	eqssd 3948	1 ⊢ ((∪ 𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝐴) → ∪ 𝐴 = 𝐵)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ⊆ wss 3899  ∪ cuni 4867

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-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3916  df-uni 4868

This theorem is used by:  elpwuni  5065  mretopd  23318  toponmre  23319  neiptopuni  23356  filunibas  24108  unidmvol  25770  unicls  34414  carsguni  34820  onintunirab  44069