Metamath Proof Explorer


Theorem rexraleqim

Description: Statement following from existence and generalization with equality. (Contributed by AV, 9-Feb-2019)

Ref Expression
Hypotheses rexraleqim.1 ⊢ x = z → ψ ↔ φ
rexraleqim.2 ⊢ z = Y → φ ↔ θ
Assertion rexraleqim ⊢ ∃ z ∈ A φ ∧ ∀ x ∈ A ψ → x = Y → θ

Proof

Step Hyp Ref Expression
1 rexraleqim.1 ⊢ x = z → ψ ↔ φ
2 rexraleqim.2 ⊢ z = Y → φ ↔ θ
3 eqeq1 ⊢ x = z → x = Y ↔ z = Y
4 1 3 imbi12d ⊢ x = z → ψ → x = Y ↔ φ → z = Y
5 4 rspcva ⊢ z ∈ A ∧ ∀ x ∈ A ψ → x = Y → φ → z = Y
6 2 biimpd ⊢ z = Y → φ → θ
7 5 6 syli ⊢ z ∈ A ∧ ∀ x ∈ A ψ → x = Y → φ → θ
8 7 impancom ⊢ z ∈ A ∧ φ → ∀ x ∈ A ψ → x = Y → θ
9 8 rexlimiva ⊢ ∃ z ∈ A φ → ∀ x ∈ A ψ → x = Y → θ
10 9 imp ⊢ ∃ z ∈ A φ ∧ ∀ x ∈ A ψ → x = Y → θ