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Proof step 51 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⊥, n3 < n0 + n, m1 < n0 + n, m1 < # buf + n0, n3 < # buf + n0, m1 < n1, n3 < n1, afs-cons(dirs, files1), ino ∈ files0, ino ∈ files1, n0 + n ≤ n1, n1 ≤ # buf + n0, n ≤ # buf ⊦ buf[n3 - n0] = ((content0.stream to n0) + (⌜buf⌝ to n) + (content0.stream from n0 + n))[n3], n = 0, n3 < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
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⊥, n0 < n1, n3 < n0 + n, m1 < n0 + n, m1 < # buf + n0, n3 < # buf + n0, m1 < n1, n3 < n1, afs-cons(dirs, files1), ino ∈ files0, ino ∈ files1, n0 ≤ n3, n0 + n ≤ n1, n1 ≤ # buf + n0, n ≤ # buf ⊦ # buf = 0, buf[n3 - n0] = ((content0.stream to n0) + (⌜buf⌝ to n) + (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 ↔ n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)