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Proof step 13 from proof in-remove-refsize0-refsize-cons (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 14:
(∅)[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 = ∅

Rule argument

pos: left 7 terms: m abort: F discard: T computed: T

Additional information

Used terms: m