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Proof step 9 from proof id-idx-lpt-free-unref (previous step)

# lpt1 = # lpt0, lpt0[n] = mklp(m, m0, lpf, m1), lpf = LP_FREE, n < # lpt0, ∀ n. n < # lpt0 ∧ lpt1[n].flags = LP_FREE → lpt1[n].ref-size = 0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# lpt1 = # lpt0, lpt0[n] = mklp(m, m0, LP_FREE, m1), n < # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags, ∀ n. n < # lpt0 ∧ lpt1[n].flags = LP_FREE → lpt1[n].ref-size = 0 ⊦ m = 0

Proof step 9 from proof id-idx-lpt-free-unref (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information