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Proof step 34 from proof remove-refsize0-subset-refed (previous step)

st[m] = mkblock(x0, lax0, 0), (∅)[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st, m ∈ st, as ⊆ dom nodes(st), ∀ adr. adr ∈ as → adr.size ≠ 0, ∀ 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 35:
(∅)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax0, 0), m ∈ st, @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, as ⊆ dom nodes(st), @(m, m0, m1) ∈ as → @(m, m0, m1).size ≠ 0, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅

Rule argument

pos: left 7 terms: @(m, m0, m1) abort: F discard: T computed: T

Additional information

Used terms: @(m, m0, m1)