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 𝑥 𝜓
rspc.2 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion rspc ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 rspc.1 𝑥 𝜓
2 rspc.2 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
3 df-ral ( ∀ 𝑥𝐵 𝜑 ↔ ∀ 𝑥 ( 𝑥𝐵𝜑 ) )
4 nfcv 𝑥 𝐴
5 nfv 𝑥 𝐴𝐵
6 5 1 nfim 𝑥 ( 𝐴𝐵𝜓 )
7 eleq1 ( 𝑥 = 𝐴 → ( 𝑥𝐵𝐴𝐵 ) )
8 7 2 imbi12d ( 𝑥 = 𝐴 → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) )
9 4 6 8 spcgf ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝑥𝐵𝜑 ) → ( 𝐴𝐵𝜓 ) ) )
10 9 pm2.43a ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝑥𝐵𝜑 ) → 𝜓 ) )
11 3 10 syl5bi ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝜑𝜓 ) )