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Proof step 9 from proof refsize-cons-append (previous step)

st[n] = mkblock(x, lax, m), n ∈ st, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m + m1 = (as | n).size + @(n, m0, m1).size, @(m0, m1) ∈ lax, @(n, m0, m1) ∈ as | n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st[n] = mkblock(x, lax, m), n ∈ st, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m = (as | n).size, @(m0, m1) ∈ lax, @(n, m0, m1) ∈ as

Rule argument

Additional information

Used simplifier rules: restrict-in: ⊦ @(m, m0, m1) ∈ as | m ↔ @(m, m0, m1) ∈ as;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ @(m, m0, m1).size = m1; (s)