Metamath Proof Explorer


Theorem cbvex2ivw

Description: One direction of cbvex2vw does not need a DV condition between x and y . Idea from relopabVD . (Contributed by SN, 27-Oct-2026)

Ref Expression
Hypothesis cbvex2ivw.1 ⊢ x = z ∧ y = w → φ → ψ
Assertion cbvex2ivw ⊢ ∃ x ∃ y φ → ∃ z ∃ w ψ

Proof

Step Hyp Ref Expression
1 cbvex2ivw.1 ⊢ x = z ∧ y = w → φ → ψ
2 ax6evr ⊢ ∃ z x = z
3 ax6evr ⊢ ∃ w y = w
4 1 ex ⊢ x = z → y = w → φ → ψ
5 4 eximdv ⊢ x = z → ∃ w y = w → ∃ w φ → ψ
6 3 5 mpi ⊢ x = z → ∃ w φ → ψ
7 19.37v ⊢ ∃ w φ → ψ ↔ φ → ∃ w ψ
8 6 7 sylib ⊢ x = z → φ → ∃ w ψ
9 2 8 eximii ⊢ ∃ z φ → ∃ w ψ
10 9 19.37iv ⊢ φ → ∃ z ∃ w ψ
11 10 exlimivv ⊢ ∃ x ∃ y φ → ∃ z ∃ w ψ