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Proof step 19 from proof id-idx-lpt-free-unref (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# lpt1 = # lpt0, lpt0[n].flags = LP_FREE, lpt1[n] = mklp(lpt0[n].ref-size, lpt0[n].size, LP_FREE, m1), n < # lpt0, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE → lpt0[n].ref-size = 0 ⊦ lpt0[n].ref-size = 0

Rule argument

Additional information

Used simplifier rules: ⊦ mklp(n2, n3, lpf, m).ref-size = n2; (s)% ⊦ mklp(n2, n3, lpf, m).size = n3; (s)% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s)