Theorem undisj1 3522

Description: The union of disjoint classes is disjoint. (Contributed by NM, 26-Sep-2004.)

Assertion

Ref	Expression
undisj1	⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)

Proof of Theorem undisj1

Step	Hyp	Ref	Expression
1		un00 3511	. 2 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
2		indir 3426	. . 3 ⊢ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶))
3	2	eqeq1i 2214	. 2 ⊢ (((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅ ↔ ((𝐴 ∩ 𝐶) ∪ (𝐵 ∩ 𝐶)) = ∅)
4	1, 3	bitr4i 187	1 ⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)

Syntax hints:   ∧ wa 104   ↔ wb 105   = wceq 1373   ∪ cun 3168   ∩ cin 3169  ∅c0 3464

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-dif 3172  df-un 3174  df-in 3176  df-ss 3183  df-nul 3465

This theorem is referenced by:  funtp  5336