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Metamath Proof Explorer

Theorem sseqtrd

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

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

Proof

Step Hyp Ref Expression
1 sseqtrd.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2 sseqtrd.2 ⊢ ( 𝜑 → 𝐵 = 𝐶 )
3 2 sseq2d ⊢ ( 𝜑 → ( 𝐴 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐶 ) )
4 1 3 mpbid ⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 )