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Metamath Proof Explorer

Theorem ssfid

Description: A subset of a finite set is finite, deduction version of ssfi . (Contributed by Glauco Siliprandi, 21-Nov-2020)

Ref Expression
Hypotheses ssfid.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
ssfid.2 ⊢ ( 𝜑 → 𝐵 ⊆ 𝐴 )
Assertion ssfid ( 𝜑 → 𝐵 ∈ Fin )

Proof

Step Hyp Ref Expression
1 ssfid.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
2 ssfid.2 ⊢ ( 𝜑 → 𝐵 ⊆ 𝐴 )
3 ssfi ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ⊆ 𝐴 ) → 𝐵 ∈ Fin )
4 1 2 3 syl2anc ⊢ ( 𝜑 → 𝐵 ∈ Fin )