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Proof step 17 from proof id-idx-freelist-cons (previous step)

# lpt1 = # lpt0, lpt1[n] = mklp(m, m0, lpf, m1), lpt0[n] = mklp(m2, m3, lpf0, m4), n < # lpt0 → m2 = m ∧ m3 = m0 ∧ lpf0 = lpf, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist ⊦ n < # lpt0 ∧ lpf = LP_FREE ↔ n ∈ freelist, dups(freelist)

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n abort: F discard: T computed: T

Additional information

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