Theorem difjust 3935

Description: Soundness justification theorem for df-dif 3936. (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		eleq1w 2816	. . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴))
2		eleq1w 2816	. . . . 5 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵))
3	2	notbid 318	. . . 4 ⊢ (𝑥 = 𝑧 → (¬ 𝑥 ∈ 𝐵 ↔ ¬ 𝑧 ∈ 𝐵))
4	1, 3	anbi12d 632	. . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵)))
5	4	cbvabv 2804	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵)}
6		eleq1w 2816	. . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
7		eleq1w 2816	. . . . 5 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
8	7	notbid 318	. . . 4 ⊢ (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝐵 ↔ ¬ 𝑦 ∈ 𝐵))
9	6, 8	anbi12d 632	. . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)))
10	9	cbvabv 2804	. 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ ¬ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)}
11	5, 10	eqtri 2757	1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)}

Syntax hints:  ¬ wn 3   ∧ wa 395   = wceq 1539   ∈ wcel 2107  {cab 2712

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808

This theorem is referenced by: (None)