Metamath Proof Explorer


Theorem rspc

Description: Restricted specialization, using implicit substitution. (Contributed by NM, 19-Apr-2005) (Revised by Mario Carneiro, 11-Oct-2016)

Ref Expression
Hypotheses rspc.1 ⊢ Ⅎ 𝑥 𝜓
rspc.2 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion rspc ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ∈ 𝐵 𝜑 → 𝜓 ) )

Proof

Step Hyp Ref Expression
1 rspc.1 ⊢ Ⅎ 𝑥 𝜓
2 rspc.2 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
3 df-ral ⊢ ( ∀ 𝑥 ∈ 𝐵 𝜑 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝜑 ) )
4 nfcv ⊢ Ⅎ 𝑥 𝐴
5 nfv ⊢ Ⅎ 𝑥 𝐴 ∈ 𝐵
6 5 1 nfim ⊢ Ⅎ 𝑥 ( 𝐴 ∈ 𝐵 → 𝜓 )
7 eleq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵 ) )
8 7 2 imbi12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 ∈ 𝐵 → 𝜑 ) ↔ ( 𝐴 ∈ 𝐵 → 𝜓 ) ) )
9 4 6 8 spcgf ⊢ ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝜑 ) → ( 𝐴 ∈ 𝐵 → 𝜓 ) ) )
10 9 pm2.43a ⊢ ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝜑 ) → 𝜓 ) )
11 3 10 biimtrid ⊢ ( 𝐴 ∈ 𝐵 → ( ∀ 𝑥 ∈ 𝐵 𝜑 → 𝜓 ) )