Metamath Proof Explorer


Theorem pssssd

Description: Deduce subclass from proper subclass. (Contributed by NM, 29-Feb-1996)

Ref Expression
Hypothesis pssssd.1 ⊢ ( 𝜑 → 𝐴 ⊊ 𝐵 )
Assertion pssssd ( 𝜑 → 𝐴 ⊆ 𝐵 )

Proof

Step Hyp Ref Expression
1 pssssd.1 ⊢ ( 𝜑 → 𝐴 ⊊ 𝐵 )
2 pssss ⊢ ( 𝐴 ⊊ 𝐵 → 𝐴 ⊆ 𝐵 )
3 1 2 syl ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )