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Proof step 11 from proof avol-replay-lpt-unchanged-write-right (previous step)

# 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[n0], n0 = n

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n0

Additional information

Used simplifier rules: ⊦ mklp(n2, n3, lpf, m).ref-size = n2; (s)% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s) Used terms: n0