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Proof step 5 from proof freelist-cons-add (previous step)

lpt[n, mklp(m0, m1, LP_FREE, m3)][n3] = mklp(m, m2, lpf0, m4), lpt[n] = mklp(n0, n1, lpf, n2), n3 = n, n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, dups(freelist), ¬ n ∈ freelist ∧ (n3 < # lpt ∧ lpf0 = LP_FREE ↔ n3 ∈ freelist ∨ n3 = n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
lpt[n] = mklp(n0, n1, lpf, n2), n < # lpt, n ∈ freelist, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, dups(freelist)

Rule argument

Additional information

Used simplifier rules: ⊦ mklp(m, m0, lpf, m1) = mklp(m2, m3, lpf0, m4) ↔ m = m2 ∧ m0 = m3 ∧ lpf = lpf0 ∧ m1 = m4; (ws)% ⊦ m < # lpt → lpt[m, lp][m] = lp; (s)