Theorem difab 4234

Description: Difference of two class abstractions. (Contributed by NM, 23-Oct-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
difab	⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)}

Proof of Theorem difab

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-clab 2716	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓))
2		sban 2083	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓))
3		df-clab 2716	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4	3	bicomi 223	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑})
5		sbn 2277	. . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓)
6		df-clab 2716	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
7	5, 6	xchbinxr 335	. . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓})
8	4, 7	anbi12i 627	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}))
9	1, 2, 8	3bitrri 298	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)})
10	9	difeqri 4059	1 ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)}

Syntax hints:  ¬ wn 3   ∧ wa 396   = wceq 1539  [wsb 2067   ∈ wcel 2106  {cab 2715   ∖ cdif 3884

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-12 2171  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890

This theorem is referenced by:  notab  4238  difrab  4242  notrab  4245