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Theorem sssymdifcl 44572
Description: The class of all subsets of a class is closed under symmetric difference. (Contributed by RP, 3-Jan-2020.)
Hypothesis
Ref Expression
ssficl.a 𝐴 = {𝑧 ∣ 𝑧 ⊆ 𝐵}
Assertion
Ref Expression
sssymdifcl ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ∈ 𝐴
Distinct variable groups:   𝑥,𝑦,𝑧   𝑦,𝐴   𝑧,𝐵
Allowed substitution hints:   𝐴(𝑥, 𝑧)   𝐵(𝑥, 𝑦)

Proof of Theorem sssymdifcl
StepHypRef Expression
1 ssficl.a . 2 𝐴 = {𝑧 ∣ 𝑧 ⊆ 𝐵}
2 vex 3455 . . . 4 𝑥 ∈ V
32difexi 5292 . . 3 (𝑥 ∖ 𝑦) ∈ V
4 vex 3455 . . . 4 𝑦 ∈ V
54difexi 5292 . . 3 (𝑦 ∖ 𝑥) ∈ V
63, 5unex 7761 . 2 ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ∈ V
7 sseq1 3956 . 2 (𝑧 = ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) → (𝑧 ⊆ 𝐵 ↔ ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ⊆ 𝐵))
8 sseq1 3956 . 2 (𝑧 = 𝑥 → (𝑧 ⊆ 𝐵 ↔ 𝑥 ⊆ 𝐵))
9 sseq1 3956 . 2 (𝑧 = 𝑦 → (𝑧 ⊆ 𝐵 ↔ 𝑦 ⊆ 𝐵))
10 ssdifss 4087 . . 3 (𝑥 ⊆ 𝐵 → (𝑥 ∖ 𝑦) ⊆ 𝐵)
11 ssdifss 4087 . . 3 (𝑦 ⊆ 𝐵 → (𝑦 ∖ 𝑥) ⊆ 𝐵)
12 unss 4136 . . . 4 (((𝑥 ∖ 𝑦) ⊆ 𝐵 ∧ (𝑦 ∖ 𝑥) ⊆ 𝐵) ↔ ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ⊆ 𝐵)
1312biimpi 219 . . 3 (((𝑥 ∖ 𝑦) ⊆ 𝐵 ∧ (𝑦 ∖ 𝑥) ⊆ 𝐵) → ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ⊆ 𝐵)
1410, 11, 13syl2an 608 . 2 ((𝑥 ⊆ 𝐵 ∧ 𝑦 ⊆ 𝐵) → ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ⊆ 𝐵)
151, 6, 7, 8, 9, 14cllem0 44566 1 ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ((𝑥 ∖ 𝑦) ∪ (𝑦 ∖ 𝑥)) ∈ 𝐴
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2739  ∀wral 3077  Vcvv 3451   ∖ cdif 3896   ∪ cun 3897   ⊆ wss 3899
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733  ax-sep 5249  ax-pr 5391  ax-un 7751
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-sn 4585  df-pr 4587  df-uni 4868
This theorem is used by: (None)
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