Metamath Proof Explorer
Description: Distribution of class substitution over conjunction, in inference form.
(Contributed by Giovanni Mascellani, 27-May-2019)
|
|
Ref |
Expression |
|
Hypotheses |
sbcani.1 |
⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜒 ) |
|
|
sbcani.2 |
⊢ ( [ 𝐴 / 𝑥 ] 𝜓 ↔ 𝜂 ) |
|
Assertion |
sbcani |
⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( 𝜒 ∧ 𝜂 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sbcani.1 |
⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝜒 ) |
2 |
|
sbcani.2 |
⊢ ( [ 𝐴 / 𝑥 ] 𝜓 ↔ 𝜂 ) |
3 |
|
sbcan |
⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ) |
4 |
1 2
|
anbi12i |
⊢ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ↔ ( 𝜒 ∧ 𝜂 ) ) |
5 |
3 4
|
bitri |
⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( 𝜒 ∧ 𝜂 ) ) |