Metamath Proof Explorer


Theorem rabeqsnd

Description: Conditions for a restricted class abstraction to be a singleton, in deduction form. (Contributed by Thierry Arnoux, 2-Dec-2021)

Ref Expression
Hypotheses rabeqsnd.0 ⊢ x = B → ψ ↔ χ
rabeqsnd.1 ⊢ φ → B ∈ A
rabeqsnd.2 ⊢ φ → χ
rabeqsnd.3 ⊢ φ ∧ x ∈ A ∧ ψ → x = B
Assertion rabeqsnd ⊢ φ → x ∈ A | ψ = B

Proof

Step Hyp Ref Expression
1 rabeqsnd.0 ⊢ x = B → ψ ↔ χ
2 rabeqsnd.1 ⊢ φ → B ∈ A
3 rabeqsnd.2 ⊢ φ → χ
4 rabeqsnd.3 ⊢ φ ∧ x ∈ A ∧ ψ → x = B
5 4 expl ⊢ φ → x ∈ A ∧ ψ → x = B
6 5 alrimiv ⊢ φ → ∀ x x ∈ A ∧ ψ → x = B
7 2 3 jca ⊢ φ → B ∈ A ∧ χ
8 7 a1d ⊢ φ → x = B → B ∈ A ∧ χ
9 8 alrimiv ⊢ φ → ∀ x x = B → B ∈ A ∧ χ
10 eleq1 ⊢ x = B → x ∈ A ↔ B ∈ A
11 10 1 anbi12d ⊢ x = B → x ∈ A ∧ ψ ↔ B ∈ A ∧ χ
12 11 pm5.74i ⊢ x = B → x ∈ A ∧ ψ ↔ x = B → B ∈ A ∧ χ
13 12 albii ⊢ ∀ x x = B → x ∈ A ∧ ψ ↔ ∀ x x = B → B ∈ A ∧ χ
14 9 13 sylibr ⊢ φ → ∀ x x = B → x ∈ A ∧ ψ
15 6 14 jca ⊢ φ → ∀ x x ∈ A ∧ ψ → x = B ∧ ∀ x x = B → x ∈ A ∧ ψ
16 albiim ⊢ ∀ x x ∈ A ∧ ψ ↔ x = B ↔ ∀ x x ∈ A ∧ ψ → x = B ∧ ∀ x x = B → x ∈ A ∧ ψ
17 15 16 sylibr ⊢ φ → ∀ x x ∈ A ∧ ψ ↔ x = B
18 rabeqsn ⊢ x ∈ A | ψ = B ↔ ∀ x x ∈ A ∧ ψ ↔ x = B
19 17 18 sylibr ⊢ φ → x ∈ A | ψ = B