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Proof step 10 from proof lpt-free-unref-write (previous step)

lpt[n] = mklp(m2, m3, lpf0, m4), n < # lpt ∧ lpf0 = LP_FREE → m2 = 0, n < # lpt ∧ lpf = LP_FREE → m = 0, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ ∀ n0. n0 < # lpt ∧ lpt[n, mklp(m, m0, lpf, m1)][n0].flags = LP_FREE → lpt[n, mklp(m, m0, lpf, m1)][n0].ref-size = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
lpt[n, mklp(m, m0, lpf, m1)][n0].flags = LP_FREE, lpt[n] = mklp(m2, m3, lpf0, m4), n0 < # lpt, n < # lpt ∧ lpf0 = LP_FREE → m2 = 0, n < # lpt ∧ lpf = LP_FREE → m = 0, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ lpt[n, mklp(m, m0, lpf, m1)][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