Metamath Proof Explorer


Theorem sseqtrrid

Description: Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011)

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

Proof

Step Hyp Ref Expression
1 sseqtrrid.1 ⊢ 𝐵 ⊆ 𝐴
2 sseqtrrid.2 ⊢ ( 𝜑 → 𝐶 = 𝐴 )
3 2 eqcomd ⊢ ( 𝜑 → 𝐴 = 𝐶 )
4 1 3 sseqtrid ⊢ ( 𝜑 → 𝐵 ⊆ 𝐶 )