Metamath Proof Explorer


Theorem ellaz

Description: Elementhood in an adjunction of a complex number to the ring of integers. Such an element is also called "extended integer" in the following. (Contributed by AV, 24-Oct-2026)

Ref Expression
Assertion ellaz Could not format assertion : No typesetting found for |- ( S e. CC -> ( A e. ( Z[]l ` S ) <-> ( A e. CC /\ E. a e. ZZ E. b e. ZZ A = ( a + ( b x. S ) ) ) ) ) with typecode |-

Proof

Step Hyp Ref Expression
1 lazval Could not format ( S e. CC -> ( Z[]l ` S ) = { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } ) : No typesetting found for |- ( S e. CC -> ( Z[]l ` S ) = { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } ) with typecode |-
2 1 eleq2d Could not format ( S e. CC -> ( A e. ( Z[]l ` S ) <-> A e. { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } ) ) : No typesetting found for |- ( S e. CC -> ( A e. ( Z[]l ` S ) <-> A e. { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } ) ) with typecode |-
3 eqeq1 ⊢ x = A → x = a + b ⁢ S ↔ A = a + b ⁢ S
4 3 2rexbidv ⊢ x = A → ∃ a ∈ ℤ ∃ b ∈ ℤ x = a + b ⁢ S ↔ ∃ a ∈ ℤ ∃ b ∈ ℤ A = a + b ⁢ S
5 eqid ⊢ x ∈ ℂ | ∃ a ∈ ℤ ∃ b ∈ ℤ x = a + b ⁢ S = x ∈ ℂ | ∃ a ∈ ℤ ∃ b ∈ ℤ x = a + b ⁢ S
6 4 5 elrab2 ⊢ A ∈ x ∈ ℂ | ∃ a ∈ ℤ ∃ b ∈ ℤ x = a + b ⁢ S ↔ A ∈ ℂ ∧ ∃ a ∈ ℤ ∃ b ∈ ℤ A = a + b ⁢ S
7 2 6 bitrdi Could not format ( S e. CC -> ( A e. ( Z[]l ` S ) <-> ( A e. CC /\ E. a e. ZZ E. b e. ZZ A = ( a + ( b x. S ) ) ) ) ) : No typesetting found for |- ( S e. CC -> ( A e. ( Z[]l ` S ) <-> ( A e. CC /\ E. a e. ZZ E. b e. ZZ A = ( a + ( b x. S ) ) ) ) ) with typecode |-