Metamath Proof Explorer


Theorem dvelimexcased

Description: Eliminate a disjoint variable condition from an existentially quantified statement using cases. (Contributed by BTernaryTau, 31-Jul-2025)

Ref Expression
Hypotheses dvelimexcased.1 ⊢ Ⅎ x φ
dvelimexcased.2 ⊢ ¬ ∀ x x = y → Ⅎ z φ
dvelimexcased.3 ⊢ φ ∧ ¬ ∀ x x = y → Ⅎ x ψ
dvelimexcased.4 ⊢ φ ∧ ¬ ∀ x x = y → Ⅎ z θ
dvelimexcased.5 ⊢ φ ∧ ¬ ∀ x x = y → z = x → ψ → θ
dvelimexcased.6 ⊢ φ ∧ ∀ x x = y → χ → θ
dvelimexcased.7 ⊢ φ → ∃ z ψ
dvelimexcased.8 ⊢ φ → ∃ x χ
Assertion dvelimexcased ⊢ φ → ∃ x θ

Proof

Step Hyp Ref Expression
1 dvelimexcased.1 ⊢ Ⅎ x φ
2 dvelimexcased.2 ⊢ ¬ ∀ x x = y → Ⅎ z φ
3 dvelimexcased.3 ⊢ φ ∧ ¬ ∀ x x = y → Ⅎ x ψ
4 dvelimexcased.4 ⊢ φ ∧ ¬ ∀ x x = y → Ⅎ z θ
5 dvelimexcased.5 ⊢ φ ∧ ¬ ∀ x x = y → z = x → ψ → θ
6 dvelimexcased.6 ⊢ φ ∧ ∀ x x = y → χ → θ
7 dvelimexcased.7 ⊢ φ → ∃ z ψ
8 dvelimexcased.8 ⊢ φ → ∃ x χ
9 nfa1 ⊢ Ⅎ x ∀ x x = y
10 1 9 nfan ⊢ Ⅎ x φ ∧ ∀ x x = y
11 10 6 eximd ⊢ φ ∧ ∀ x x = y → ∃ x χ → ∃ x θ
12 11 ex ⊢ φ → ∀ x x = y → ∃ x χ → ∃ x θ
13 8 12 mpid ⊢ φ → ∀ x x = y → ∃ x θ
14 nfv ⊢ Ⅎ z ¬ ∀ x x = y
15 14 2 nfan1c ⊢ Ⅎ z φ ∧ ¬ ∀ x x = y
16 nfna1 ⊢ Ⅎ x ¬ ∀ x x = y
17 1 16 nfan ⊢ Ⅎ x φ ∧ ¬ ∀ x x = y
18 15 17 3 4 5 cbvex1v ⊢ φ ∧ ¬ ∀ x x = y → ∃ z ψ → ∃ x θ
19 18 ex ⊢ φ → ¬ ∀ x x = y → ∃ z ψ → ∃ x θ
20 7 19 mpid ⊢ φ → ¬ ∀ x x = y → ∃ x θ
21 13 20 pm2.61d ⊢ φ → ∃ x θ