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Metamath Proof Explorer

Theorem sseq12i

Description: An equality inference for the subclass relationship. (Contributed by NM, 31-May-1999) (Proof shortened by Eric Schmidt, 26-Jan-2007)

Ref Expression
Hypotheses sseq1i.1 ⊢ 𝐴 = 𝐵
sseq12i.2 ⊢ 𝐶 = 𝐷
Assertion sseq12i ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷 )

Proof

Step Hyp Ref Expression
1 sseq1i.1 ⊢ 𝐴 = 𝐵
2 sseq12i.2 ⊢ 𝐶 = 𝐷
3 sseq12 ⊢ ( ( 𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ) → ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷 ) )
4 1 2 3 mp2an ⊢ ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷 )