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Proof step 3 from proof avol-replay-lpt-unchanged-extract-right (previous step)

# avol0 = # avol1, # lpt = # avol1, n < # avol1, ∀ n. n < # avol1 ∧ (n ∈ nx ∨ lpt[n].ref-size ≠ 0 ∨ lpt[n].flags = LP_FREE) → avol0[n] = avol1[n], n ∈ nx ∨ lpt[n].ref-size ≠ 0 ∨ lpt[n].flags = LP_FREE ⊦ avol1[n] = avol0[n]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# lpt = # avol1, # avol0 = # avol1, lpt[n] = mklp(m, m0, lpf, m1), n < # avol1, ∀ n. n < # avol1 ∧ (n ∈ nx ∨ lpt[n].ref-size ≠ 0 ∨ lpt[n].flags = LP_FREE) → avol0[n] = avol1[n], n ∈ nx ∨ m ≠ 0 ∨ lpf = LP_FREE ⊦ avol1[n] = avol0[n]

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[n], m, m0, lpf]

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[n], m, m0, lpf