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Proof step 1 from proof lpt-free-unref-replay-lpt (conclusion)

n ≤ # lpt ⊦ lpt-free-unref(replay-lpt(lpt, nx, n)) ↔ lpt-free-unref(lpt)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n ≤ # lpt ⊦ (∀ n0. n0 < # lpt ∧ replay-lpt(lpt, nx, n)[n0].flags = LP_FREE → replay-lpt(lpt, nx, n)[n0].ref-size = 0) ↔ (∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0)

Rule argument

Additional information

Used simplifier rules: lpt-free-unref: ⊦ lpt-free-unref(lpt) ↔ (∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0);%replay-lpt-size: n ≤ # lpt ⊦ # replay-lpt(lpt, nx, n) = # lpt;% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)