Theorem ssdif0im 3329

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 819	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		eldif 2993	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 635	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
4	3	alimi 1385	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
5		dfss2 2999	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6		eq0 3284	. 2 ⊢ ((𝐴 ∖ 𝐵) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
7	4, 5, 6	3imtr4i 199	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∖ 𝐵) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102  ∀wal 1283   = wceq 1285   ∈ wcel 1434   ∖ cdif 2981   ⊆ wss 2984  ∅c0 3269

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2614  df-dif 2986  df-in 2990  df-ss 2997  df-nul 3270

This theorem is referenced by:  vdif0im  3330  difrab0eqim  3331  difid  3333  difin0  3338