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Proof step 6 from proof avol-replay-lpt-unchanged-write-right (previous step)

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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n

Additional information

Used simplifier rules: ⊦ m < # avol → avol[m, aleb][m] = aleb; (s)% ⊦ mklp(n2, n3, lpf, m).ref-size = n2; (s)% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s) Used terms: n