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Proof step 6 from proof freelist-cons-alloc (previous step)

lpt[n0] = mklp(m, m3, lpf0, m4), ∀ n0. n0 < # lpt ∧ lpt[n0].flags = LP_FREE ↔ n0 = n ∨ n0 ∈ freelist ⊦ n0 < # lpt ∧ lpf0 = LP_FREE ↔ n0 ∈ freelist, lpf = LP_FREE, n0 = n, dups(freelist), n ∈ freelist

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: n0

Additional information

Used simplifier rules: ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s) Used terms: n0