Metamath Proof Explorer
Description: Deduction for inequality. (Contributed by NM, 25Oct1999) (Proof
shortened by Wolf Lammen, 19Nov2019)


Ref 
Expression 

Hypothesis 
neeq1d.1 
⊢ ( 𝜑 → 𝐴 = 𝐵 ) 

Assertion 
neeq2d 
⊢ ( 𝜑 → ( 𝐶 ≠ 𝐴 ↔ 𝐶 ≠ 𝐵 ) ) 
Proof
Step 
Hyp 
Ref 
Expression 
1 

neeq1d.1 
⊢ ( 𝜑 → 𝐴 = 𝐵 ) 
2 
1

eqeq2d 
⊢ ( 𝜑 → ( 𝐶 = 𝐴 ↔ 𝐶 = 𝐵 ) ) 
3 
2

necon3bid 
⊢ ( 𝜑 → ( 𝐶 ≠ 𝐴 ↔ 𝐶 ≠ 𝐵 ) ) 