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Proof step 4 from proof refsize-cons-append (previous step)

st[n] = mkblock(x, lax, m), st[n, mkblock(x + a ', lax + @(m0, m1) ', m + m1)][n0] = mkblock(x0, lax0, m2), n0 ∈ st[n, mkblock(x + a ', lax + @(m0, m1) ', m + m1)], n ∈ st, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m2 = ((as ++ @(n, m0, m1)) | n0).size, @(m0, m1) ∈ lax

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ≠ n0 → st[n0, ablock][n] = st[n]
Proof step 6:
st[n, mkblock(x + a ', lax + @(m0, m1) ', m + m1)][n0] = mkblock(x0, lax0, m2), st[n] = mkblock(x, lax, m), n0 = n, n ∈ st, n0 ∈ st[n, mkblock(x + a ', lax + @(m0, m1) ', m + m1)], as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m2 = ((as ++ @(n, m0, m1)) | n0).size, @(m0, m1) ∈ lax
Proof step 11:
st[n0] = mkblock(x0, lax0, m2), st[n] = mkblock(x, lax, m), n ∈ st, n0 ∈ st[n, mkblock(x + a ', lax + @(m0, m1) ', m + m1)], as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m2 = ((as ++ @(n, m0, m1)) | n0).size, n0 = n, @(m0, m1) ∈ lax

Rule argument

spec: store, inst: ablock-store-nat-store name: At-other seq: ⊦ n ≠ n0 → st[n0, ablock][n] = st[n] subst: [n, ablock, n0, st] → [n0, mkblock(x + a ', lax + @(m0, m1) ', m + m1), n, st] flags: <all paths>

Additional information

Used terms: n0, mkblock(x + a ', lax + @(m0, m1) ', m + m1), n, st