Metamath Proof Explorer
Description: An equivalence law for equality. (Contributed by NM, 1-Aug-1993) (Proof
shortened by Wolf Lammen, 10-Dec-2017)
|
|
Ref |
Expression |
|
Assertion |
equequ1 |
⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑧 ↔ 𝑦 = 𝑧 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ax7 |
⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑧 → 𝑦 = 𝑧 ) ) |
2 |
|
equtr |
⊢ ( 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → 𝑥 = 𝑧 ) ) |
3 |
1 2
|
impbid |
⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑧 ↔ 𝑦 = 𝑧 ) ) |