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Proof step 234 from proof inv-rename-new-reparent (previous step)

m1 + 1 = fs[ri[inodekey(m)]].size, n0 + (boolvar2 ⊃ 1;0) = fs[ri[inodekey(m)]].nsubdirs, fs[ri[dentrykey(m, str)]] = dentrynode(key, m3), fs[ri[inodekey(m3)]] = inodenode(key0, md1, boolvar0, m5, m9, m10), fs[ri[inodekey(m)]].directory, fs[ri[inodekey(m0)]].directory, fo.inodes?, ro.inodes?, file-node-cons(fs, ri), log-cons(log, fs), orphans-cons(fs, ri, ro), replay-cons-strong(fs, ri, ro, fi, fo, log), ri-inj(fi), ri-inj(ri), ubifs-cons(fs, ri, ro, fi, fo, log), dentrykey(m, str) ∈ ri, inodekey(m3) ∈ ri, inodekey(m0) ∈ ri, inodekey(m) ∈ ri, ri[dentrykey(m, str)] ∈ fs, ri[inodekey(m3)] ∈ fs, ⌊err0⌋ , fs[ri[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO, fs[ri[inodekey(m0)]].nlink ≠ 0 ∨ m0 = ROOT_INO, ¬ boolvar0 ∧ ¬ boolvar2 ∨ boolvar0 ∧ boolvar2 ∧ m4 = m10 ⊦ ⟪skip⟫ (ubifs-cons(fs, ri, ro, fi, fo, log) ∧ replay-cons-strong(fs, ri, ro, fi, fo, log) ∧ file-node-cons(fs, ri) ∧ orphans-cons(fs, ri, ro)), m0 = m, inodekey(m0) ∈ ro, dentrykey(m0, str0) ∈ ri, inodekey(m) ∈ ro

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

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