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Mirrors > Home > MPE Home > Th. List > eqssd | Structured version Visualization version GIF version |
Description: Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 27-Jun-2004.) |
Ref | Expression |
---|---|
eqssd.1 | ⊢ (𝜑 → 𝐴 ⊆ 𝐵) |
eqssd.2 | ⊢ (𝜑 → 𝐵 ⊆ 𝐴) |
Ref | Expression |
---|---|
eqssd | ⊢ (𝜑 → 𝐴 = 𝐵) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqssd.1 | . 2 ⊢ (𝜑 → 𝐴 ⊆ 𝐵) | |
2 | eqssd.2 | . 2 ⊢ (𝜑 → 𝐵 ⊆ 𝐴) | |
3 | eqss 3937 | . 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)) | |
4 | 1, 2, 3 | sylanbrc 583 | 1 ⊢ (𝜑 → 𝐴 = 𝐵) |
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