Theorem difjust 3122

Description: Soundness justification theorem for df-dif 3123. (Contributed by Rodolfo Medina, 27-Apr-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)

Assertion

Ref	Expression
difjust	⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)}

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑦,𝐴   𝑦,𝐵

Proof of Theorem difjust

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq1 2233	. . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴))
2		eleq1 2233	. . . . 5 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵))
3	2	notbid 662	. . . 4 ⊢ (𝑥 = 𝑧 → (¬ 𝑥 ∈ 𝐵 ↔ ¬ 𝑧 ∈ 𝐵))
4	1, 3	anbi12d 470	. . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵)))
5	4	cbvabv 2295	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵)}
6		eleq1 2233	. . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
7		eleq1 2233	. . . . 5 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
8	7	notbid 662	. . . 4 ⊢ (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝐵 ↔ ¬ 𝑦 ∈ 𝐵))
9	6, 8	anbi12d 470	. . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)))
10	9	cbvabv 2295	. 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)}
11	5, 10	eqtri 2191	1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)}

Syntax hints:  ¬ wn 3   ∧ wa 103   = wceq 1348   ∈ wcel 2141  {cab 2156

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166

This theorem is referenced by: (None)