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Proof step 21 from proof rename-precond-abs (previous step)

fs[ri[inodekey(m3)]].nlink = 0, new_dent = negdentry(str0), fs[ri[inodekey(m3)]].directory, fs[ri[inodekey(m4)]].directory, orphans(fs, ri).inodes?, new_dent.negdentry?, afs-cons(abs-dirs(fs, ri), abs-files(fs, ri)), file-node-cons(fs, ri), log-cons(log, fs), replay-cons-strong(fs, ri, orphans(fs, ri), fi, fo, log), ri-inj(ri), ri-inj(fi), ubifs-cons(fs, ri, orphans(fs, ri), fi, fo, log), inodekey(m4) ∈ ri, inodekey(m3) ∈ ri, dentrykey(m4, str) ∈ ri, inodekey(fs[ri[dentrykey(m4, str)]].ino) ∈ ri, m3 = m4 → md = md0, fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].directory ∧ m8 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].size ∧ m7 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].nsubdirs ∧ boolvar0 ∨ ¬ fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].directory ∧ m7 = 0 ∧ ¬ boolvar0, links(m4, abs-dirs(fs, ri)) ≠ ∅ ∨ m4 = ROOT_INO ⊦ # links(m3, abs-dirs(fs, ri)) = 0, links(m3, abs-dirs(fs, ri)) = ∅, m3 = ROOT_INO, dentrykey(m3, str0) ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
fs[ri[inodekey(m3)]].nlink = 0, fs[ri[inodekey(m4)]].directory, fs[ri[inodekey(m3)]].directory, orphans(fs, ri).inodes?, afs-cons(abs-dirs(fs, ri), abs-files(fs, ri)), file-node-cons(fs, ri), log-cons(log, fs), 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), inodekey(fs[ri[dentrykey(m4, str)]].ino) ∈ ri, dentrykey(m4, str) ∈ ri, inodekey(m3) ∈ ri, inodekey(m4) ∈ ri, m3 = m4 → md = md0, links(m4, abs-dirs(fs, ri)) ≠ ∅ ∨ m4 = ROOT_INO, fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].directory ∧ m8 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].size ∧ m7 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].nsubdirs ∧ boolvar0 ∨ ¬ fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].directory ∧ m7 = 0 ∧ ¬ boolvar0 ⊦ # links(m3, abs-dirs(fs, ri)) = 0, links(m3, abs-dirs(fs, ri)) = ∅, m3 = ROOT_INO, dentrykey(m3, str0) ∈ ri

Rule argument

Additional information

Used simplifier rules: ⊦ negdentry(str).negdentry?; (s)