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Proof step 7 from proof refsize-cons-write (previous step)

st[n] = mkblock(x0, lax0, m0), ∀ n. n ∈ st → st[n].ref-size = (as | n).size, ∀ n0. n0 ∈ st[n, mkblock(x, lax, m)] → st[n, mkblock(x, lax, m)][n0].ref-size = (as | n0).size, ¬ n ∈ st ∨ m0 = (as | n).size ⊦ m = (as | n).size

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n

Additional information

Used simplifier rules: ⊦ n ∈ st[n, ablock]; (s)% ⊦ st[n, ablock][n] = ablock; (s)% ⊦ mkblock(x, lax, m).ref-size = m; (s) Used terms: n