Theorem ssdifsn

Description: Subset of a set with an element removed. (Contributed by Emmett Weisz, 7-Jul-2021) (Proof shortened by JJ, 31-May-2022)

	Ref	Expression
Assertion	ssdifsn	⊢ ( 𝐴 ⊆ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		difss2	⊢ ( 𝐴 ⊆ ( 𝐵 ∖ { 𝐶 } ) → 𝐴 ⊆ 𝐵 )
2		reldisj	⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝐴 ∩ { 𝐶 } ) = ∅ ↔ 𝐴 ⊆ ( 𝐵 ∖ { 𝐶 } ) ) )
3	2	bicomd	⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐴 ⊆ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ∩ { 𝐶 } ) = ∅ ) )
4	1 3	biadanii	⊢ ( 𝐴 ⊆ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ⊆ 𝐵 ∧ ( 𝐴 ∩ { 𝐶 } ) = ∅ ) )
5		disjsn	⊢ ( ( 𝐴 ∩ { 𝐶 } ) = ∅ ↔ ¬ 𝐶 ∈ 𝐴 )
6	5	anbi2i	⊢ ( ( 𝐴 ⊆ 𝐵 ∧ ( 𝐴 ∩ { 𝐶 } ) = ∅ ) ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) )
7	4 6	bitri	⊢ ( 𝐴 ⊆ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) )