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Proof step 4 from proof freelist-cons-first-free (previous step)

lpt[n] = mklp(m, m0, lpf, m1), ∀ n0. n0 < # lpt ∧ lpt[n0].flags = LP_FREE ↔ n0 = n ∨ n0 ∈ freelist ⊦ dups(freelist), n ∈ freelist, n < # lpt ∧ lpf = LP_FREE

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s) Used terms: n