Proof step 10 from proof refsize-cons-append (previous step)
st[n] = mkblock(x, lax, m), n ∈ st, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size
⊦ m = (as | n).size, @(m0, m1) ∈ lax, @(n, m0, m1) ∈ as
User applied rule all left
And closed the goal!
Rule argument
pos: left 4 terms: n abort: F discard: T computed: T
Additional information
Used simplifier rules:
⊦ mkblock(x, lax, m).ref-size = m; (s)
Used terms: n