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Proof step 149 from proof afs-cons-rename-file-dentry (previous step)

# links(entries0[str], dirs0) = m15 + 1, dirs0[entries0[str]] = mkdir(m16, md9, m17, m18, entries2), dirs0[m] = mkdir(m6 + 1, md6, # links(m, dirs0), m8, entries0), files0[entries0[str0]] = mkfile(m1, md3, # links(entries0[str0], dirs0), content), files0[entries0[str]] = mkfile(m5, md0, m15 + 1, content1), afs-cons(dirs0, files0), ROOT_INO ∈ dirs0, m ∈ dirs0, str ∈ entries0, entries0[str0] ∈ files0, str0 ∈ entries0, entries0[str] ∈ files0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries0[str0] = entries0[str], m15 = #(links(entries0[str], dirs0) -- mklink(m, str)), mkdentry(str, entries0[str]) = mkdentry(str0, entries0[str0]), entries0[str0] ∈ dirs0, entries0[str] ∈ dirs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: mklink(m, str) ∈ links(entries0[str], dirs0)

Additional information

Used simplifier rules: ⊦ ¬ alink ∈ ls → ls -- alink = ls; (s)%⊦ alink ∈ links(ino, dirs) ↔ alink.pino ∈ dirs ∧ alink.name ∈ dirs[alink.pino].entries ∧ ino = dirs[alink.pino].entries[alink.name]; (s)% ⊦ mklink(m, str).pino = m; (s)% ⊦ mklink(m, str).name = str; (s)% ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ alink ∈ ls → #(ls -- alink) = # ls - 1; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s) Used formulas: mklink(m, str) ∈ links(entries0[str], dirs0)