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Proof step 5 from proof lebrefs-append-none (previous step)

∀ n. (q0 + eq-entry(n, none) ').lebrefs = q0.lebrefs ⊦ (q0 + eq-entry(n, none) ').lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅) = q0.lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
lref = none, ∀ n. (q0 + eq-entry(n, none) ').lebrefs = q0.lebrefs ⊦ (q0 + eq-entry(n, none) ').lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅) = q0.lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅)
Proof step 8:
∀ n. (q0 + eq-entry(n, none) ').lebrefs = q0.lebrefs ⊦ lref = none, (q0 + eq-entry(n, none) ').lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅) = q0.lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅)

Rule argument

fma: lref = none

Additional information

Used formulas: lref = none