Description: An element not in a set can be removed without affecting the set. (Contributed by NM, 16-Mar-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)
Ref | Expression | ||
---|---|---|---|
Assertion | difsn | ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( 𝐵 ∖ { 𝐴 } ) = 𝐵 ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eldifsn | ⊢ ( 𝑥 ∈ ( 𝐵 ∖ { 𝐴 } ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝑥 ≠ 𝐴 ) ) | |
2 | simpl | ⊢ ( ( 𝑥 ∈ 𝐵 ∧ 𝑥 ≠ 𝐴 ) → 𝑥 ∈ 𝐵 ) | |
3 | nelelne | ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( 𝑥 ∈ 𝐵 → 𝑥 ≠ 𝐴 ) ) | |
4 | 3 | ancld | ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐵 ∧ 𝑥 ≠ 𝐴 ) ) ) |
5 | 2 4 | impbid2 | ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( ( 𝑥 ∈ 𝐵 ∧ 𝑥 ≠ 𝐴 ) ↔ 𝑥 ∈ 𝐵 ) ) |
6 | 1 5 | bitrid | ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( 𝑥 ∈ ( 𝐵 ∖ { 𝐴 } ) ↔ 𝑥 ∈ 𝐵 ) ) |
7 | 6 | eqrdv | ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( 𝐵 ∖ { 𝐴 } ) = 𝐵 ) |