Theorem ssdif0im 3529

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 690	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		eldif 3179	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 679	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
4	3	alimi 1479	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
5		ssalel 3185	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6		eq0 3483	. 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 1371   = wceq 1373   ∈ wcel 2177   ∖ cdif 3167   ⊆ wss 3170  ∅c0 3464

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 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-dif 3172  df-in 3176  df-ss 3183  df-nul 3465

This theorem is referenced by:  vdif0im  3530  difrab0eqim  3531  difid  3533  difin0  3538