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Proof step 7 from proof lpt-free-unref-write (previous step)

lpt[n] = mklp(m2, m3, lpf0, m4), n < # lpt ∧ lpf0 = LP_FREE → m2 = 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) ↔ n < # lpt ∧ lpf = LP_FREE → m = 0

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 8:
lpt[n] = mklp(m2, m3, lpf0, m4), lpf = LP_FREE, n < # lpt, n < # lpt ∧ lpf0 = LP_FREE → m2 = 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 ⊦ m = 0
Proof step 10:
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

Rule argument

pos: right 1

Proof step 7 from proof lpt-free-unref-write (previous step)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

Additional information