Theorem inssdif0 4329

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 4321	. 2 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ ((𝐴 ∩ 𝐵) ∖ 𝐶) = ∅)
2		indif2 4234	. . 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 1570   ∖ cdif 3902   ∩ cin 3904   ⊆ wss 3905  ∅c0 4286

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-dif 3908  df-in 3912  df-ss 3922  df-nul 4287

This theorem is used by:  inindif  4331  disjdifg  4432  inf3lem3  9595  ssfin4  10298  isnrm2  23524  1stccnp  23628  llycmpkgen2  23716  ufileu  24085  fclscf  24191  flimfnfcls  24194  opnbnd  36864  ttcwf2  37064  diophrw  43518  setindtr  43779