Theorem difab 3523

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 2340	. . 3 ⊢ (y ∈ {x ∣ (φ ∧ ¬ ψ)} ↔ [y / x](φ ∧ ¬ ψ))
2		sban 2069	. . 3 ⊢ ([y / x](φ ∧ ¬ ψ) ↔ ([y / x]φ ∧ [y / x] ¬ ψ))
3		df-clab 2340	. . . . 5 ⊢ (y ∈ {x ∣ φ} ↔ [y / x]φ)
4	3	bicomi 193	. . . 4 ⊢ ([y / x]φ ↔ y ∈ {x ∣ φ})
5		sbn 2062	. . . . 5 ⊢ ([y / x] ¬ ψ ↔ ¬ [y / x]ψ)
6		df-clab 2340	. . . . 5 ⊢ (y ∈ {x ∣ ψ} ↔ [y / x]ψ)
7	5, 6	xchbinxr 302	. . . 4 ⊢ ([y / x] ¬ ψ ↔ ¬ y ∈ {x ∣ ψ})
8	4, 7	anbi12i 678	. . 3 ⊢ (([y / x]φ ∧ [y / x] ¬ ψ) ↔ (y ∈ {x ∣ φ} ∧ ¬ y ∈ {x ∣ ψ}))
9	1, 2, 8	3bitrri 263	. 2 ⊢ ((y ∈ {x ∣ φ} ∧ ¬ y ∈ {x ∣ ψ}) ↔ y ∈ {x ∣ (φ ∧ ¬ ψ)})
10	9	difeqri 3387	1 ⊢ ({x ∣ φ} ∖ {x ∣ ψ}) = {x ∣ (φ ∧ ¬ ψ)}

Syntax hints:  ¬ wn 3   ∧ wa 358   = wceq 1642  [wsb 1648   ∈ wcel 1710  {cab 2339   ∖ cdif 3206

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215

This theorem is referenced by:  notab  3525  difrab  3529  notrab  3532