Metamath Proof Explorer


Theorem rspct

Description: A closed version of rspc . (Contributed by Andrew Salmon, 6-Jun-2011)

Ref Expression
Hypothesis rspct.1 𝑥 𝜓
Assertion rspct ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝜑𝜓 ) ) )

Proof

Step Hyp Ref Expression
1 rspct.1 𝑥 𝜓
2 df-ral ( ∀ 𝑥𝐵 𝜑 ↔ ∀ 𝑥 ( 𝑥𝐵𝜑 ) )
3 eleq1 ( 𝑥 = 𝐴 → ( 𝑥𝐵𝐴𝐵 ) )
4 3 adantr ( ( 𝑥 = 𝐴 ∧ ( 𝜑𝜓 ) ) → ( 𝑥𝐵𝐴𝐵 ) )
5 simpr ( ( 𝑥 = 𝐴 ∧ ( 𝜑𝜓 ) ) → ( 𝜑𝜓 ) )
6 4 5 imbi12d ( ( 𝑥 = 𝐴 ∧ ( 𝜑𝜓 ) ) → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) )
7 6 ex ( 𝑥 = 𝐴 → ( ( 𝜑𝜓 ) → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) ) )
8 7 a2i ( ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝑥 = 𝐴 → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) ) )
9 8 alimi ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ∀ 𝑥 ( 𝑥 = 𝐴 → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) ) )
10 nfv 𝑥 𝐴𝐵
11 10 1 nfim 𝑥 ( 𝐴𝐵𝜓 )
12 nfcv 𝑥 𝐴
13 11 12 spcgft ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( ( 𝑥𝐵𝜑 ) ↔ ( 𝐴𝐵𝜓 ) ) ) → ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝑥𝐵𝜑 ) → ( 𝐴𝐵𝜓 ) ) ) )
14 9 13 syl ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝐴𝐵 → ( ∀ 𝑥 ( 𝑥𝐵𝜑 ) → ( 𝐴𝐵𝜓 ) ) ) )
15 2 14 syl7bi ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝜑 → ( 𝐴𝐵𝜓 ) ) ) )
16 15 com34 ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝐴𝐵 → ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝜑𝜓 ) ) ) )
17 16 pm2.43d ( ∀ 𝑥 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) ) → ( 𝐴𝐵 → ( ∀ 𝑥𝐵 𝜑𝜓 ) ) )