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Proof step 73 from proof inv-write-inode (previous step)

fs0[ri[inodekey(m)]].nlink = 0 ∧ m ≠ ROOT_INO ↔ inodekey(m) ∈ ro, fs0[ri[inodekey(m)]].inode?, fo.inodes?, ro.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(fi), ri-inj(ri), ubifs-cons(fs0, ri, ro, fi, fo, log0), ri[inodekey(m)] ∈ fs0, inodekey(m) ∈ ri, inodekey(ino) ∈ ri, ¬ fs0[ri[inodekey(ino)]].directory → fs0[ri[inodekey(ino)]].nsubdirs = 0, ¬ fs0[ri[inodekey(m)]].directory ∧ ¬ boolvar0 ∨ fs0[ri[inodekey(m)]].directory ∧ boolvar0 ∧ m2 = fs0[ri[inodekey(m)]].size ⊦ (adr1 = ri[inodekey(ino)] ⊃ inodenode(inodekey(m), md, boolvar0, fs0[ri[inodekey(m)]].nlink, fs0[ri[inodekey(m)]].nsubdirs, m2) ;fs0[ri[inodekey(ino)]] ) .nsubdirs = 0, m = ino, (adr1 = ri[inodekey(ino)] ⊃ inodenode(inodekey(m), md, boolvar0, fs0[ri[inodekey(m)]].nlink, fs0[ri[inodekey(m)]].nsubdirs, m2) ;fs0[ri[inodekey(ino)]] ) .directory, adr1 ∈ fs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ inodenode(key, md, boolvar, m, m0, m1).nsubdirs = m0; (s)% ⊦ key ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri[inodekey(ino)]