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Proof step 20 from proof refsize-cons-write (previous step)

st[n] = mkblock(x0, lax0, m0), st[n0] = mkblock(x1, lax1, m1), n ∈ st, n0 ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m0 = (as | n).size, m1 = (as | n0).size, n0 = n

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n0

Additional information

Used simplifier rules: ⊦ mkblock(x, lax, m).ref-size = m; (s) Used terms: n0