Theorem difab

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

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

Proof

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