Theorem inssdif0 4325

Description: Intersection, subclass, and difference relationship. (Contributed by NM, 27-Oct-1996.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof shortened by Wolf Lammen, 30-Sep-2014.) (Proof shortened by BJ, 18-Jul-2026.)

Assertion

Ref	Expression
inssdif0	⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅)

Proof of Theorem inssdif0

Step	Hyp	Ref	Expression
1		ssdif0 4317	. 2 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ ((𝐴 ∩ 𝐵) ∖ 𝐶) = ∅)
2		indif2 4229	. . 3 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
3	2	eqeq1i 2768	. 2 ⊢ ((𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅ ↔ ((𝐴 ∩ 𝐵) ∖ 𝐶) = ∅)
4	1, 3	bitr4i 281	1 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1569   ⊆ wss 3900   ∖ cdif 4048   ∩ cin 4092  ∅c0 4281

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-nul 4282

This theorem is used by:  inindif  4327  disjdifg  4428  inf3lem3  9644  ssfin4  10401  isnrm2  23691  1stccnp  23796  llycmpkgen2  23884  ufileu  24253  fclscf  24359  flimfnfcls  24362  opnbnd  37246  ttcwf2  37446  diophrw  43916  setindtr  44177