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Proof step 238 from proof inv-rename-new-reparent (previous step)

fs[ri[dentrykey(m, str)]] = dentrynode(key, m1), fs[ri[dentrykey(m, str)]].dentry?, fs[ri[inodekey(m)]].directory, fs[ri[inodekey(m0)]].directory, fs[ri[inodekey(m1)]].inode?, fo.inodes?, ro.inodes?, fs[ri[inodekey(m)]].nsubdirs < (boolvar2 ⊃ 1;0), 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(m1) ∈ ri, inodekey(m0) ∈ ri, inodekey(m) ∈ ri, ri[dentrykey(m, str)] ∈ fs, ri[inodekey(m1)] ∈ fs, 1 ≤ fs[ri[inodekey(m)]].size, boolvar2 → 1 ≤ fs[ri[inodekey(m)]].nsubdirs, fs[ri[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO, fs[ri[inodekey(m0)]].nlink ≠ 0 ∨ m0 = ROOT_INO, ¬ fs[ri[inodekey(m1)]].directory ∧ ¬ boolvar2 ∨ fs[ri[inodekey(m1)]].directory ∧ boolvar2 ∧ m4 = fs[ri[inodekey(m1)]].size ⊦ fs[ri[inodekey(m)]].size = 0, m0 = m, inodekey(m0) ∈ ro, dentrykey(m0, str0) ∈ ri, inodekey(m) ∈ ro

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ dentrynode(key, m).dentry?; (s)