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

lpt[n, mklp(m0, m1, lpf, m2)][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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m ≠ n → lpt[n, lp][m] = lpt[m]; (s)