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Metamath Proof Explorer

Theorem eqimssd

Description: Equality implies inclusion, deduction version. (Contributed by SN, 6-Nov-2024)

Ref Expression
Hypothesis eqimssd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion eqimssd ( 𝜑 → 𝐴 ⊆ 𝐵 )

Proof

Step Hyp Ref Expression
1 eqimssd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 ssid ⊢ 𝐵 ⊆ 𝐵
3 1 2 eqsstrdi ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )