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Proof step 55 from proof inv-renam (previous step)

fs[ri[inodekey(m4)]].size ≠ 0, fs[ri[inodekey(m4)]].directory, fo.inodes?, ro.inodes?, file-node-cons(fs0, ri0), file-node-cons(fs, ri), log-cons(log0, fs0), log-cons(log, fs), orphans-cons(fs0, ri0, ro), orphans-cons(fs, ri, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), replay-cons-strong(fs, ri, ro, fi, fo, log), ri-inj(ri0), ri-inj(fi), ri-inj(ri), ubifs-cons(fs0, ri0, ro, fi, fo, log0), ubifs-cons(fs, ri, ro, fi, fo, log), inodekey(m4) ∈ ri, inodekey(fs[ri[dentrykey(m4, str)]].ino) ∈ ri, dentrykey(m4, str) ∈ ri, 1 ≤ fs[ri[inodekey(m4)]].size, ¬ dentrykey(m4, str0) ∈ ri, ¬ inodekey(m4) ∈ ro, boolvar → 1 ≤ fs[ri[inodekey(m4)]].nsubdirs, ¬ fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].directory ∧ ¬ boolvar ∧ m0 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].nlink ∧ m1 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].nsubdirs ∨ fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].directory ∧ boolvar ∧ m0 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].nlink ∧ m1 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].nsubdirs ∧ m2 = fs[ri[inodekey(fs[ri[dentrykey(m4, str)]].ino)]].size, fs[ri[inodekey(m4)]].nlink ≠ 0 ∨ m4 = ROOT_INO ⊦ ⟪new_parent_inode := old_parent_inode0⟫ true

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

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