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Proof step 5 from proof prepost-create-freelist (previous step)

freelist = [], n = 0 ⊦ ⟪while n < # lpt do {/* (prepost-create-freelist: invariant n ≤ # lpt ∧ freelist-cons(subarray(lpt, 0, n), freelist) wfbound # lpt - n) */; if lpt[n].flags = LP_FREE then freelist := freelist + n; n := n + 1}⟫ freelist-cons(lpt, freelist)

Heuristic annotation applied rule annotation right

And obtained the following new goals

Proof step 6:
freelist = [], n = 0 ⊦ n ≤ # lpt ∧ freelist-cons(subarray(lpt, 0, n), freelist)
Proof step 9:
freelist0 = [], n = n0, n1 = 0, n < # lpt, freelist-cons(subarray(lpt, 0, n), freelist), n ≤ # lpt ⊦ ⟪if lpt[n].flags = LP_FREE then freelist := freelist + n; n := n + 1⟫ ((n ≤ # lpt ∧ freelist-cons(subarray(lpt, 0, n), freelist)) ∧ # lpt - n < # lpt - n0)

Rule argument

rule: invariant right arg: bound: # lpt - n inv: n ≤ # lpt ∧ freelist-cons(subarray(lpt, 0, n), freelist) fmapos: right 1 pos: right 1

Additional information

Used simplifier rules: ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ n ≠ n0 → (n < n0 ↔ ¬ n0 < n); (ss)% ⊦ m < n → ¬ n < m; (s)%# lpt = n ⊦ subarray(lpt, 0, n) = lpt; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) Used formulas: n ≤ # lpt ∧ freelist-cons(subarray(lpt, 0, n), freelist), # lpt - n = # lpt - n