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Proof step 35 from proof remove-refsize0-refsize-cons (previous step)

# x = # lax1 + # lax0 + 1, st[m] = mkblock(x, lax1 + @(m0, 0) ' + lax0, 0), @(m, m0, 0) ∈ as, m ∈ st, as ⊆ dom nodes(st), ∀ n. ¬ # lax1 + # lax0 < n → (lax1 + @(m0, 0) ' + lax0)[n].size = x[n].flashsize ⊦ # as = 0, as = ∅, dups(lax1 + lax0), dups(lax1), dups(lax0), @(m0, 0) ∈ lax1, @(m0, 0) ∈ lax0, dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: # lax1

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (lax + ladr ' + lax0)[# lax] = ladr; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ @(m, m0).size = m0; (s)% ⊦ a.flashsize ≠ 0; (s) Used terms: # lax1