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Proof step 33 from proof refine-lookup (previous step)

fs[ri[dentrykey(p_ino, old_dent.name)]] = dentrynode(key, m), fs[ri[inodekey(p_ino)]].directory, fo.inodes?, orphans(fs, ri).inodes?, afs-cons(abs-dirs(fs, ri), abs-files(fs, ri)), file-node-cons(fs, ri), log-cons(log, fs), orphans-cons(fs, ri, orphans(fs, ri)), replay-cons-strong(fs, ri, orphans(fs, ri), fi, fo, log), ri-inj(fi), ri-inj(ri), ubifs-cons(fs, ri, orphans(fs, ri), fi, fo, log), old_dent.name ∈ abs-children(p_ino, fs, ri), inodekey(p_ino) ∈ ri, ri[dentrykey(p_ino, old_dent.name)] ∈ fs, dentrykey(p_ino, old_dent.name) ∈ ri, asynced → boolvar1, links(p_ino, abs-dirs(fs, ri)) ≠ ∅ ∨ p_ino = ROOT_INO, fs[ri[inodekey(p_ino)]].nlink ≠ 0 ∨ p_ino = ROOT_INO ⊦ ⟨err := ESUCCESS⟩ ((asynced → boolvar1) ∧ mkdentry(old_dent.name, m) = mkdentry(old_dent.name, abs-children(p_ino, fs, ri)[old_dent.name]) ∧ ESUCCESS = err), ⟨fail#(; ; err)⟩ ((asynced → boolvar1) ∧ mkdentry(old_dent.name, m) = old_dent ∧ ESUCCESS = err)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ str ∈ abs-children(ino, fs, ri) ↔ dentrykey(ino, str) ∈ ri; (s)%⊦ dentrykey(ino, str) ∈ ri → abs-children(ino, fs, ri)[str] = fs[ri[dentrykey(ino, str)]].ino; (s)% ⊦ dentrynode(key, m).ino = m; (s)