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Proof step 6 from proof in-remove-refsize0-refsize-cons (previous step)

adr = @(m, m0, m1), inv(st), adr ∈ as, adr ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m ∈ remove-refsize0(st) ∧ @(m0, m1) ∈ remove-refsize0(st)[m].ladrlist

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
inv(st), @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m ∈ remove-refsize0(st) ∧ @(m0, m1) ∈ remove-refsize0(st)[m].ladrlist

Proof step 6 from proof in-remove-refsize0-refsize-cons (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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