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Proof step 31 from proof remove-refsize0-refsize-cons (previous step)

# x = # lax, st[m] = mkblock(x, lax, 0), @(m, m0, 0) ∈ as, m ∈ st, @(m, m0, 0).lnum ∈ st, @(@(m, m0, 0).pos, @(m, m0, 0).size) ∈ st[@(m, m0, 0).lnum].ladrlist, as ⊆ dom nodes(st), ∀ n. n < # lax → lax[n].size = x[n].flashsize ⊦ # as = 0, as = ∅, dups(lax), dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
# x = # lax, st[m] = mkblock(x, lax, 0), @(m, m0, 0) ∈ as, m ∈ st, @(m0, 0) ∈ lax, as ⊆ dom nodes(st), ∀ n. n < # lax → lax[n].size = x[n].flashsize ⊦ # as = 0, as = ∅, dups(lax), dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

Rule argument

Additional information

Used simplifier rules: ⊦ mkblock(x, lax, m).ladrlist = lax; (s)% ⊦ @(m, m0, m1).lnum = m; (s)% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ @(m, m0, m1).pos = m0; (s)