Theorem ssdif0im 3280

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 784	. . . 4 ⊢ ((x ∈ A → x ∈ B) → ¬ (x ∈ A ∧ ¬ x ∈ B))
2		eldif 2921	. . . 4 ⊢ (x ∈ (A ∖ B) ↔ (x ∈ A ∧ ¬ x ∈ B))
3	1, 2	sylnibr 601	. . 3 ⊢ ((x ∈ A → x ∈ B) → ¬ x ∈ (A ∖ B))
4	3	alimi 1341	. 2 ⊢ (∀x(x ∈ A → x ∈ B) → ∀x ¬ x ∈ (A ∖ B))
5		dfss2 2928	. 2 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
6		eq0 3233	. 2 ⊢ ((A ∖ B) = ∅ ↔ ∀x ¬ x ∈ (A ∖ B))
7	4, 5, 6	3imtr4i 190	1 ⊢ (A ⊆ B → (A ∖ B) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97  ∀wal 1240   = wceq 1242   ∈ wcel 1390   ∖ cdif 2908   ⊆ wss 2911  ∅c0 3218

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-in 2918  df-ss 2925  df-nul 3219

This theorem is referenced by:  vdif0im  3281  difrab0eqim  3282  difid  3286  difin0  3291