Theorem inssdif0 4322

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 4314	. 2 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ ((𝐴 ∩ 𝐵) ∖ 𝐶) = ∅)
2		indif2 4227	. . 3 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
3	2	eqeq1i 2766	. 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 3896   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279

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 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3453  df-dif 3902  df-in 3906  df-ss 3916  df-nul 4280

This theorem is used by:  inindif  4324  disjdifg  4425  inf3lem3  9631  ssfin4  10388  isnrm2  23676  1stccnp  23781  llycmpkgen2  23869  ufileu  24238  fclscf  24344  flimfnfcls  24347  opnbnd  37113  ttcwf2  37313  diophrw  43769  setindtr  44030