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Metamath Proof Explorer

Theorem eqsstrd

Description: Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004)

Ref Expression
Hypotheses eqsstrd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
eqsstrd.2 ⊢ ( 𝜑 → 𝐵 ⊆ 𝐶 )
Assertion eqsstrd ( 𝜑 → 𝐴 ⊆ 𝐶 )

Proof

Step Hyp Ref Expression
1 eqsstrd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 eqsstrd.2 ⊢ ( 𝜑 → 𝐵 ⊆ 𝐶 )
3 1 sseq1d ⊢ ( 𝜑 → ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶 ) )
4 2 3 mpbird ⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 )