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Proof step 8 from proof avol-replay-lpt-unchanged-write-right (previous step)

# avol0 = # avol1, # lpt = # avol1, lpt[n] = mklp(m, m0, lpf, m1), n < # avol1, n0 < # avol1, ∀ n. n < # avol1 ∧ (n ∈ nx ∨ lpt[n].ref-size ≠ 0 ∨ lpt[n].flags = LP_FREE) → avol0[n] = avol1[n], aleb = avol0[n] ∨ ¬ n ∈ nx ∧ m = 0 ∧ lpf ≠ LP_FREE, n0 ∈ nx ∨ lpt[n0].ref-size ≠ 0 ∨ lpt[n0].flags = LP_FREE ⊦ avol0[n0] = avol1[n, aleb][n0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
# lpt = # avol1, # avol0 = # avol1, lpt[n] = mklp(m, m0, lpf, m1), lpt[n0] = mklp(m2, m3, lpf0, m4), n0 < # avol1, n < # avol1, ∀ n. n < # avol1 ∧ (n ∈ nx ∨ lpt[n].ref-size ≠ 0 ∨ lpt[n].flags = LP_FREE) → avol0[n] = avol1[n], n0 ∈ nx ∨ m2 ≠ 0 ∨ lpf0 = LP_FREE, aleb = avol0[n] ∨ ¬ n ∈ nx ∧ m = 0 ∧ lpf ≠ LP_FREE ⊦ avol0[n0] = avol1[n, aleb][n0]

Rule argument

spec: lprops name: elim seq: ⊦ ((m = lp.ref-size ∧ m0 = lp.size) ∧ lpf = lp.flags) ∧ m1 = lp.gcheapidx ↔ lp = mklp(m, m0, lpf, m1) subst: [lp, m, m0, lpf] → [lpt[n0], m1, m2, lpf0]

Additional information

Used simplifier rules: ⊦ ((m = lp.ref-size ∧ m0 = lp.size) ∧ lpf = lp.flags) ∧ m1 = lp.gcheapidx ↔ lp = mklp(m, m0, lpf, m1); (elim) Used terms: lpt[n0], m1, m2, lpf0