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Proof step 9 from proof remove-refsize0-refsize-cons (previous step)

inv(st), @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, as ⊆ dom nodes(st), m ∈ st → st[m].ref-size = (as | m).size ⊦ # as = 0, as = ∅, @(m, m0, m1) ∈ remove-refsize0(st), dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st[m].ref-size = (as | m).size, inv(st), @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, m ∈ st, as ⊆ dom nodes(st) ⊦ # as = 0, as = ∅, @(m, m0, m1) ∈ remove-refsize0(st), dom nodes(st) ⊆ dom nodes(remove-refsize0(st))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ st → ¬ @(n, n0, n1) ∈ st; (s)