Metamath Proof Explorer
Description: An equality theorem for substitution. (Contributed by NM, 6-Oct-2004)
(Proof shortened by Andrew Salmon, 21-Jun-2011)
|
|
Ref |
Expression |
|
Assertion |
sbequ12r |
⊢ ( 𝑥 = 𝑦 → ( [ 𝑥 / 𝑦 ] 𝜑 ↔ 𝜑 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sbequ12 |
⊢ ( 𝑦 = 𝑥 → ( 𝜑 ↔ [ 𝑥 / 𝑦 ] 𝜑 ) ) |
2 |
1
|
bicomd |
⊢ ( 𝑦 = 𝑥 → ( [ 𝑥 / 𝑦 ] 𝜑 ↔ 𝜑 ) ) |
3 |
2
|
equcoms |
⊢ ( 𝑥 = 𝑦 → ( [ 𝑥 / 𝑦 ] 𝜑 ↔ 𝜑 ) ) |