Metamath Proof Explorer
Description: Equality theorem for class substitution. (Contributed by Mario
Carneiro, 9-Feb-2017) (Revised by NM, 30-Jun-2018)
|
|
Ref |
Expression |
|
Hypothesis |
sbceq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
Assertion |
sbceq1d |
⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐵 / 𝑥 ] 𝜓 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sbceq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
dfsbcq |
⊢ ( 𝐴 = 𝐵 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐵 / 𝑥 ] 𝜓 ) ) |
3 |
1 2
|
syl |
⊢ ( 𝜑 → ( [ 𝐴 / 𝑥 ] 𝜓 ↔ [ 𝐵 / 𝑥 ] 𝜓 ) ) |