Theorem ssdif0im 3366

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 826	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		eldif 3022	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 640	. . 3 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
4	3	alimi 1396	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
5		dfss2 3028	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6		eq0 3320	. 2 ⊢ ((𝐴 ∖ 𝐵) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∖ 𝐵))
7	4, 5, 6	3imtr4i 200	1 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∖ 𝐵) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103  ∀wal 1294   = wceq 1296   ∈ wcel 1445   ∖ cdif 3010   ⊆ wss 3013  ∅c0 3302

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-v 2635  df-dif 3015  df-in 3019  df-ss 3026  df-nul 3303

This theorem is referenced by:  vdif0im  3367  difrab0eqim  3368  difid  3370  difin0  3375