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Proof step 2 from proof lpt-free-unref-write (previous step)

lpt-free-unref(lpt) ⊦ lpt-free-unref(lpt[n, lp]) ↔ n < # lpt ∧ lp.flags = LP_FREE → lp.ref-size = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ (∀ n0. n0 < # lpt ∧ lpt[n, lp][n0].flags = LP_FREE → lpt[n, lp][n0].ref-size = 0) ↔ n < # lpt ∧ lp.flags = LP_FREE → lp.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);% ⊦ # lpt[n, lp] = # lpt; (s)