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Proof step 37 from proof prepost-create-freelist (previous step)

freelist0 = [], n = n0, n1 = 0, n < # lpt, freelist-cons(subarray(lpt, 0, n), freelist), n ≤ # lpt ⊦ lpt[n].flags = LP_FREE, (n + 1 ≤ # lpt ∧ freelist-cons(subarray(lpt, 0, n + 1), freelist)) ∧ # lpt - n + 1 < # lpt - n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
n0 < # lpt, freelist-cons(subarray(lpt, 0, n0), freelist) ⊦ lpt[n0].flags = LP_FREE, freelist-cons(subarray(lpt, 0, n0 + 1), freelist)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ n < m → (m - n) + n = m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → m ≤ n; (s)