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Proof step 4 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, dups(freelist), n ∈ freelist

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
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

Rule argument

fma: n0 = n

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)%m < # lpt ⊦ lpt[m, lp][m] = lp; (s)%⊦ mklp(m, m0, lpf, m1) = mklp(m2, m3, lpf0, m4) ↔ m = m2 ∧ m0 = m3 ∧ lpf = lpf0 ∧ m1 = m4; (ws) Used formulas: n0 = n