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Proof step 72 from proof write-inv-post (previous step)

files1 -- ino = files0 -- ino, files1[ino] = mkfile(m1, md1, # links(ino, dirs), content1), files0[ino] = mkfile(m1, md1, n2, content0), content0.stream from m1 = σ⊥, n4 = n3 + n, n0 < m1, n4 + n0 < m1, afs-cons(dirs, files1), ino ∈ files0, ino ∈ files1, n ≤ n4, n0 + n ≤ n1, n1 ≤ # buf + n0, n ≤ # buf, n0 + n ≤ m1 ⊦ getbyte(n4 + n0, content0) = (content0.stream from n0 + n)[n3], n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
files1 -- ino = files0 -- ino, files0[ino] = mkfile(m1, md1, n2, content0), files1[ino] = mkfile(m1, md1, # links(ino, dirs), content1), content0.stream from m1 = σ⊥, (n3 + n) + n0 < m1, n0 < m1, afs-cons(dirs, files1), ino ∈ files1, ino ∈ files0, n0 + n ≤ m1, n ≤ # buf, n1 ≤ # buf + n0, n0 + n ≤ n1 ⊦ getbyte((n3 + n) + n0, content0) = (content0.stream from n0 + n)[n3], n = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)