Metamath Proof Explorer
Description: If in a context x is not free in ps and ch , then it is not
free in ( ps /\ ch ) . (Contributed by Mario Carneiro, 7-Oct-2016)
|
|
Ref |
Expression |
|
Hypotheses |
nfand.1 |
⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 ) |
|
|
nfand.2 |
⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 ) |
|
Assertion |
nfand |
⊢ ( 𝜑 → Ⅎ 𝑥 ( 𝜓 ∧ 𝜒 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
nfand.1 |
⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 ) |
2 |
|
nfand.2 |
⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 ) |
3 |
|
df-an |
⊢ ( ( 𝜓 ∧ 𝜒 ) ↔ ¬ ( 𝜓 → ¬ 𝜒 ) ) |
4 |
2
|
nfnd |
⊢ ( 𝜑 → Ⅎ 𝑥 ¬ 𝜒 ) |
5 |
1 4
|
nfimd |
⊢ ( 𝜑 → Ⅎ 𝑥 ( 𝜓 → ¬ 𝜒 ) ) |
6 |
5
|
nfnd |
⊢ ( 𝜑 → Ⅎ 𝑥 ¬ ( 𝜓 → ¬ 𝜒 ) ) |
7 |
3 6
|
nfxfrd |
⊢ ( 𝜑 → Ⅎ 𝑥 ( 𝜓 ∧ 𝜒 ) ) |