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Proof step 3 from proof in-vl-append0 (previous step)

⊦ (∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, none) ∈ q) ∨ none = lref0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
none = lref0 ⊦ (∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, none) ∈ q) ∨ none = lref0
Proof step 6:
⊦ (∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, none) ∈ q) ∨ none = lref0, none = lref0

Rule argument

fma: none = lref0

Additional information

Used formulas: none = lref0