Theorem disjr 3574

Description: Two ways of saying that two classes are disjoint. (Contributed by Jeff Madsen, 19-Jun-2011.)

Assertion

Ref	Expression
disjr	⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem disjr

Step	Hyp	Ref	Expression
1		incom 3421	. . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	eqeq1i 2246	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅)
3		disj 3573	. 2 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴)
4	2, 3	bitri 184	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 105   = wceq 1402   ∈ wcel 2209  ∀wral 2528   ∩ cin 3219  ∅c0 3520

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-v 2823  df-dif 3222  df-in 3226  df-nul 3521

This theorem is used by: (None)