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Proof step 10 from proof refsize-cons-alloc (
previous step
)
st[n0] = mkblock(x, lax, m), as | n = ∅, n0 ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m = (as | n0).size, n ∈ st
User applied rule all left
And closed the goal!
Rule argument
pos: left 4 terms: n0 abort: F discard: T computed: T
Additional information
Used simplifier rules: ⊦ mkblock(x, lax, m).ref-size = m; (s) Used terms: n0