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Proof step 3 from proof lpt-free-unref-cleanup-lpt (previous step)

∀ n. n < # lpt ∧ lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n].flags = LP_FREE → lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n].ref-size = 0 ⊦ ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic