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Proof step 179 from proof inv-rename-new-keep-parent (previous step)

fs0[ri0[dentrykey(m, str)]] = dentrynode(key0, m0), adr = adr2, fs = fs0[adr1, dentrynode(dentrykey(m, str), 0)][adr2, dentrynode(dentrykey(m, str0), m0)] [adr3, inodenode(inodekey(m), md, true, fs0[ri0[inodekey(m)]].nlink, fs0[ri0[inodekey(m)]].nsubdirs, fs0[ri0[inodekey(m)]].size) ], key = dentrykey(m, str0), log = ((log0 + adr1 ') + adr2 ') + adr3 ', ri = ri0 -- dentrykey(m, str), fs0[ri0[inodekey(m)]].directory, fs0[ri0[inodekey(m)]].inode?, fo.inodes?, ro.inodes?, file-node-cons(fs0, ri0), log-cons(log0, fs0), orphans-cons(fs0, ri0, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), ri-inj(fi), ri-inj(ri0), ubifs-cons(fs0, ri0, ro, fi, fo, log0), adr ∈ fs, dentrykey(m, str) ∈ ri0, inodekey(m0) ∈ ri0, ri0[dentrykey(m, str)] ∈ fs0, inodekey(m) ∈ ri0, ri0[inodekey(m)] ∈ fs0, 1 ≤ fs0[ri0[inodekey(m)]].size, boolvar2 → 1 ≤ fs0[ri0[inodekey(m)]].nsubdirs, fs0[ri0[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO, ¬ fs0[ri0[inodekey(m0)]].directory ∧ ¬ boolvar2 ∧ m1 = fs0[ri0[inodekey(m0)]].nlink ∧ m2 = fs0[ri0[inodekey(m0)]].nsubdirs ∨ fs0[ri0[inodekey(m0)]].directory ∧ boolvar2 ∧ m1 = fs0[ri0[inodekey(m0)]].nlink ∧ m2 = fs0[ri0[inodekey(m0)]].nsubdirs ∧ m3 = fs0[ri0[inodekey(m0)]].size, fs0[ri0[inodekey(m)]].trunc? ∨ fs0[ri0[inodekey(m)]].inode? ⊦ fs0[ri0[inodekey(m)]].size = 0, adr1 = adr3, adr1 = adr2, adr2 = adr3, inodekey(m) ∈ ro, dentrykey(m, str0) ∈ ri0, adr1 ∈ fs0, adr2 ∈ fs0, adr3 ∈ fs0, adr ∈ fs, adr ∈ ri

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

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