Metamath Proof Explorer
Description: Restricted specialization, using implicit substitution. (Contributed by Stanislas Polu, 9-Mar-2020)
|
|
Ref |
Expression |
|
Hypotheses |
rspcdv2.1 |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
|
|
rspcdv2.2 |
⊢ ( 𝜑 → 𝐴 ∈ 𝐵 ) |
|
|
rspcdv2.3 |
⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐵 𝜓 ) |
|
Assertion |
rspcdv2 |
⊢ ( 𝜑 → 𝜒 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
rspcdv2.1 |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
2 |
|
rspcdv2.2 |
⊢ ( 𝜑 → 𝐴 ∈ 𝐵 ) |
3 |
|
rspcdv2.3 |
⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐵 𝜓 ) |
4 |
2 1
|
rspcdv |
⊢ ( 𝜑 → ( ∀ 𝑥 ∈ 𝐵 𝜓 → 𝜒 ) ) |
5 |
3 4
|
mpd |
⊢ ( 𝜑 → 𝜒 ) |