Metamath Proof Explorer
Description: Membership in a set with an element removed : deduction version.
(Contributed by Thierry Arnoux, 4-May-2025)
|
|
Ref |
Expression |
|
Hypotheses |
eldifsnd.1 |
⊢ ( 𝜑 → 𝐴 ∈ 𝐵 ) |
|
|
eldifsnd.2 |
⊢ ( 𝜑 → 𝐴 ≠ 𝐶 ) |
|
Assertion |
eldifsnd |
⊢ ( 𝜑 → 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
eldifsnd.1 |
⊢ ( 𝜑 → 𝐴 ∈ 𝐵 ) |
2 |
|
eldifsnd.2 |
⊢ ( 𝜑 → 𝐴 ≠ 𝐶 ) |
3 |
|
eldifsn |
⊢ ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ) ) |
4 |
1 2 3
|
sylanbrc |
⊢ ( 𝜑 → 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ) |