Metamath Proof Explorer


Theorem elabd2

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab . (Contributed by Gino Giotto, 12-Oct-2024) (Revised by BJ, 16-Oct-2024)

Ref Expression
Hypotheses elabd2.ex ( 𝜑𝐴𝑉 )
elabd2.eq ( 𝜑𝐵 = { 𝑥𝜓 } )
elabd2.is ( ( 𝜑𝑥 = 𝐴 ) → ( 𝜓𝜒 ) )
Assertion elabd2 ( 𝜑 → ( 𝐴𝐵𝜒 ) )

Proof

Step Hyp Ref Expression
1 elabd2.ex ( 𝜑𝐴𝑉 )
2 elabd2.eq ( 𝜑𝐵 = { 𝑥𝜓 } )
3 elabd2.is ( ( 𝜑𝑥 = 𝐴 ) → ( 𝜓𝜒 ) )
4 2 eleq2d ( 𝜑 → ( 𝐴𝐵𝐴 ∈ { 𝑥𝜓 } ) )
5 elab6g ( 𝐴𝑉 → ( 𝐴 ∈ { 𝑥𝜓 } ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜓 ) ) )
6 4 5 sylan9bb ( ( 𝜑𝐴𝑉 ) → ( 𝐴𝐵 ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜓 ) ) )
7 elisset ( 𝐴𝑉 → ∃ 𝑥 𝑥 = 𝐴 )
8 3 pm5.74da ( 𝜑 → ( ( 𝑥 = 𝐴𝜓 ) ↔ ( 𝑥 = 𝐴𝜒 ) ) )
9 8 albidv ( 𝜑 → ( ∀ 𝑥 ( 𝑥 = 𝐴𝜓 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜒 ) ) )
10 19.23v ( ∀ 𝑥 ( 𝑥 = 𝐴𝜒 ) ↔ ( ∃ 𝑥 𝑥 = 𝐴𝜒 ) )
11 9 10 bitrdi ( 𝜑 → ( ∀ 𝑥 ( 𝑥 = 𝐴𝜓 ) ↔ ( ∃ 𝑥 𝑥 = 𝐴𝜒 ) ) )
12 pm5.5 ( ∃ 𝑥 𝑥 = 𝐴 → ( ( ∃ 𝑥 𝑥 = 𝐴𝜒 ) ↔ 𝜒 ) )
13 11 12 sylan9bb ( ( 𝜑 ∧ ∃ 𝑥 𝑥 = 𝐴 ) → ( ∀ 𝑥 ( 𝑥 = 𝐴𝜓 ) ↔ 𝜒 ) )
14 7 13 sylan2 ( ( 𝜑𝐴𝑉 ) → ( ∀ 𝑥 ( 𝑥 = 𝐴𝜓 ) ↔ 𝜒 ) )
15 6 14 bitrd ( ( 𝜑𝐴𝑉 ) → ( 𝐴𝐵𝜒 ) )
16 1 15 mpdan ( 𝜑 → ( 𝐴𝐵𝜒 ) )