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Description: There is exactly one element in a singleton. Exercise 2 of TakeutiZaring p. 15. This variation requires only that B , rather than A , be a set. (Contributed by NM, 28-Oct-2003)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | elsn2g | ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ { 𝐵 } ↔ 𝐴 = 𝐵 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | elsni | ⊢ ( 𝐴 ∈ { 𝐵 } → 𝐴 = 𝐵 ) | |
| 2 | snidg | ⊢ ( 𝐵 ∈ 𝑉 → 𝐵 ∈ { 𝐵 } ) | |
| 3 | eleq1 | ⊢ ( 𝐴 = 𝐵 → ( 𝐴 ∈ { 𝐵 } ↔ 𝐵 ∈ { 𝐵 } ) ) | |
| 4 | 2 3 | syl5ibrcom | ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 = 𝐵 → 𝐴 ∈ { 𝐵 } ) ) |
| 5 | 1 4 | impbid2 | ⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ { 𝐵 } ↔ 𝐴 = 𝐵 ) ) |