Theorem ssdif0im 3489

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 688	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		eldif 3140	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 677	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
4	3	alimi 1455	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
5		dfss2 3146	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6		eq0 3443	. 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 1351   = wceq 1353   ∈ wcel 2148   ∖ cdif 3128   ⊆ wss 3131  ∅c0 3424

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-dif 3133  df-in 3137  df-ss 3144  df-nul 3425

This theorem is referenced by:  vdif0im  3490  difrab0eqim  3491  difid  3493  difin0  3498