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Proof step 78 from proof inv-mkdir (previous step)

fs1[ri1[dentrykey(ino, str0)]] = dentrynode(dentrykey(ino, str0), m0), fs1[ri1[inodekey(ino)]].directory, fs1[ri1[inodekey(m)]].directory, fs1[ri1[inodekey(m)]].inode?, ro.inodes?, fo.inodes?, file-node-cons(fs1, ri1), log-cons(log0, fs1), orphans-cons(fs1, ri1, ro), replay-cons-strong(fs1, ri1, ro, fi, fo, log0), ri-inj(ri1), ri-inj(fi), ubifs-cons(fs1, ri1, ro, fi, fo, log0), inodekey(ino) ∈ ri1, inodekey(m0) ∈ ri1, dentrykey(ino, str0) ∈ ri1, inodekey(m) ∈ ri1, ri1[inodekey(m)] ∈ fs1, ri1[dentrykey(ino, str0)] ∈ fs1, fs1[ri1[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO ⊦ 0 = m, adr1 = adr2, adr1 = adr3, adr2 = ri1[inodekey(ino)], adr2 = adr3, adr3 = ri1[inodekey(ino)], ino = 0, m = ino, n = 0, (adr1 = ri1[inodekey(ino)] ⊃ inodenode(inodekey(m), md0, true, fs1[ri1[inodekey(m)]].nlink, fs1[ri1[inodekey(m)]].nsubdirs + 1, fs1[ri1[inodekey(m)]].size + 1) ;fs1[ri1[inodekey(ino)]] ) .directory, inodekey(ino) ∈ ro, inodekey(m0) ∈ ro, adr3 ∈ fs1, adr2 ∈ fs1, adr1 ∈ fs1, inodekey(n) ∈ ri1, inodekey(n) ∈ fo, dentrykey(m, str) ∈ ri1, inodekey(m) ∈ ro

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri1[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ dentrykey(ino, str) ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → fs[ri[inodekey(ino)]].directory; (ss)% ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]; (s)% ⊦ key ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri1[inodekey(ino)]