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Proof step 14 from proof in-remove-refsize0-refsize-cons (previous step)

(∅)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax0, 0), inv(st), m ∈ st, @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st, m ∈ st → st[m].ref-size = (as | m).size ⊦ # as = 0, as = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 = (as | m).size, (∅)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax0, 0), inv(st), m ∈ st, @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st ⊦ # as = 0, as = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ mkblock(x, lax, m).ref-size = m; (s)