Theorem ssdif0im 3556

Description: Subclass implies empty difference. One direction of Exercise 7 of [TakeutiZaring] p. 22. In classical logic this would be an equivalence. (Contributed by Jim Kingdon, 2-Aug-2018.)

Assertion

Ref	Expression
ssdif0im	⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∖ 𝐵) = ∅)

Proof of Theorem ssdif0im

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		imanim 692	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		eldif 3206	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 681	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
4	3	alimi 1501	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
5		ssalel 3212	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6		eq0 3510	. 2 ⊢ ((𝐴 ∖ 𝐵) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
7	4, 5, 6	3imtr4i 201	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∖ 𝐵) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104  ∀wal 1393   = wceq 1395   ∈ wcel 2200   ∖ cdif 3194   ⊆ wss 3197  ∅c0 3491

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 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-dif 3199  df-in 3203  df-ss 3210  df-nul 3492

This theorem is referenced by:  vdif0im  3557  difrab0eqim  3558  difid  3560  difin0  3565