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Metamath Proof Explorer

Theorem negeqd

Description: Equality deduction for negatives. (Contributed by NM, 14-May-1999)

Ref Expression
Hypothesis negeqd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion negeqd ( 𝜑 → - 𝐴 = - 𝐵 )

Proof

Step Hyp Ref Expression
1 negeqd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 negeq ⊢ ( 𝐴 = 𝐵 → - 𝐴 = - 𝐵 )
3 1 2 syl ⊢ ( 𝜑 → - 𝐴 = - 𝐵 )