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Proof step 3 from proof cleanup-lpt-n-size (previous step)

n0 +1 ≤ # lpt, ∀ lpt. n0 ≤ # lpt → # cleanup-lpt(lpt, n0) = # lpt ⊦ # cleanup-lpt(lpt, n0 +1) = # lpt

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < # lpt, ∀ lpt. n0 ≤ # lpt → # cleanup-lpt(lpt, n0) = # lpt ⊦ # (lpt[n0].ref-size = 0 ∧ lpt[n0].flags ≠ LP_FREE ⊃ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, lpt[n0].gcheapidx)];cleanup-lpt(lpt, n0)) = # lpt

Rule argument

Additional information

Used simplifier rules: cleanup-lpt-rec: n < # lpt ⊦ cleanup-lpt(lpt, n + 1) = (lpt[n].ref-size = 0 ∧ lpt[n].flags ≠ LP_FREE ⊃ cleanup-lpt(lpt, n)[n, mklp(0, 0, LP_FREE, lpt[n].gcheapidx)] ;cleanup-lpt(lpt, n) );% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m +1 = m + 1; (s)