Theorem difab 3500

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 2225	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓))
2		sban 2015	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓))
3		df-clab 2225	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4	3	bicomi 132	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑})
5		sbn 2012	. . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓)
6		df-clab 2225	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
7	5, 6	xchbinxr 694	. . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓})
8	4, 7	anbi12i 464	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥] ¬ 𝜓) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}))
9	1, 2, 8	3bitrri 207	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ ¬ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)})
10	9	difeqri 3349	1 ⊢ ({𝑥 ∣ 𝜑} ∖ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ ¬ 𝜓)}

Syntax hints:  ¬ wn 3   ∧ wa 104   = wceq 1402  [wsb 1815   ∈ wcel 2209  {cab 2224   ∖ cdif 3217

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222

This theorem is referenced by:  notab  3503  difrab  3507  notrab  3510  imadiflem  5458  imadif  5459