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Proof step 11 from proof lpt-free-unref-replay-lpt (previous step)

n ≤ # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ ∀ n0. n0 < # lpt ∧ replay-lpt(lpt, nx, n)[n0].flags = LP_FREE → replay-lpt(lpt, nx, n)[n0].ref-size = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
replay-lpt(lpt, nx, n)[n0].flags = LP_FREE, n0 < # lpt, n ≤ # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ replay-lpt(lpt, nx, n)[n0].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