Metamath Proof Explorer
Description: Rule of specialization, using implicit substitution. Analogous to
rspcdv . (Contributed by David Moews, 1-May-2017)
|
|
Ref |
Expression |
|
Hypotheses |
spcimdv.1 |
⊢ ( 𝜑 → 𝐴 ∈ 𝐵 ) |
|
|
spcdv.2 |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
|
Assertion |
spcdv |
⊢ ( 𝜑 → ( ∀ 𝑥 𝜓 → 𝜒 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
spcimdv.1 |
⊢ ( 𝜑 → 𝐴 ∈ 𝐵 ) |
2 |
|
spcdv.2 |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
3 |
2
|
biimpd |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 → 𝜒 ) ) |
4 |
1 3
|
spcimdv |
⊢ ( 𝜑 → ( ∀ 𝑥 𝜓 → 𝜒 ) ) |