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Proof step 25 from proof remove-refsize0-subset-refed (previous step)

st[m] = mkblock(x0, lax0, m3), st[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ remove-refsize0(st), @(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, st[m].ref-size = 0, st[@(m, m0, m1)] = x0[pos(@(m0, m1), lax0)], as = ∅, m2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
st[m] = mkblock(x, lax, m2), m ∈ st, @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ remove-refsize0(st), as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size, ∀ adr. adr ∈ as → adr.size ≠ 0 ⊦ # as = 0, st[@(m, m0, m1)] = x[pos(@(m0, m1), lax)], as = ∅, m2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ mkblock(x, lax, m).ref-size = m; (s)%⊦ mkblock(x, lax, m) = mkblock(x0, lax0, m0) ↔ x = x0 ∧ lax = lax0 ∧ m = m0; (ws)