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Proof step 52 from proof inv-write-inode (previous step)

fs0[ri[dentrykey(ino, str)]] = dentrynode(dentrykey(ino, str), m0), fs0[ri[inodekey(ino)]].directory, fs0[ri[inodekey(m)]].directory, fs0[ri[inodekey(m)]].inode?, ro.inodes?, fo.inodes?, file-node-cons(fs0, ri), log-cons(log0, fs0), orphans-cons(fs0, ri, ro), replay-cons-strong(fs0, ri, ro, fi, fo, log0), ri-inj(ri), ri-inj(fi), ubifs-cons(fs0, ri, ro, fi, fo, log0), inodekey(m) ∈ ri, ri[inodekey(m)] ∈ fs0, ri[dentrykey(ino, str)] ∈ fs0, dentrykey(ino, str) ∈ ri, inodekey(m0) ∈ ri, inodekey(ino) ∈ ri ⊦ 0 = m, adr1 = (m = ino ⊃ adr1;ri[inodekey(ino)]), ino = 0, fs0[(m = ino ⊃ adr1;ri[inodekey(ino)])].directory, inodekey(m) ∈ ro, adr1 ∈ fs0, inodekey(m0) ∈ ro, inodekey(ino) ∈ ro, fs0[ri[inodekey(m)]].nlink = 0 ∧ m ≠ ROOT_INO

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: m = ino

Additional information

Used simplifier rules: ⊦ dentrykey(ino, str) ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → fs[ri[inodekey(ino)]].directory; (ss) Used formulas: m = ino