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Proof step 292 from proof inv-rename-overwrite-keep-parent (previous step)

(adr1 = ri0[inodekey(ino)] ⊃ dentrynode(dentrykey(m, str), 0);fs0[ri0[inodekey(ino)]]).nlink = 0, ¬ boolvar1 ↔ ¬ boolvar0, fs0[ri0[dentrykey(m, str0)]] = dentrynode(key4, m1), fs0[ri0[dentrykey(m, str)]] = dentrynode(key0, m2), fs0[ri0[inodekey(m2)]] = inodenode(key1, md0, boolvar0, m5, m9, m10), fs0[ri0[inodekey(m1)]] = inodenode(key3, md3, boolvar1, 1, m11, m12), links(m1, fs0, ri0) = ∅ ++ dentrykey(m, str0), 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-cons(ri0, fs0), ri-inj(fi), ri-inj(ri0), ubifs-cons(fs0, ri0, ro, fi, fo, log0), dentrykey(m, str) ∈ ri0, inodekey(m2) ∈ ri0, dentrykey(m, str0) ∈ ri0, inodekey(m1) ∈ ri0, ri0[dentrykey(m, str)] ∈ fs0, ri0[inodekey(m2)] ∈ fs0, inodekey(m) ∈ ri0, ri0[dentrykey(m, str0)] ∈ fs0, ri0[inodekey(m1)] ∈ fs0, ri0[inodekey(m)] ∈ fs0, inodekey(ino) ∈ ri0, boolvar1 → dentrykeys(inodekey(m1), ri0) = ∅, m4 ≠ 0 ∨ m = ROOT_INO, ¬ boolvar0 ∧ ¬ boolvar3 ∨ boolvar0 ∧ boolvar3 ∧ m3 = m10, ¬ boolvar0 ∧ ¬ boolvar2 ∨ boolvar1 ∧ boolvar2 ∧ m6 = m12 ⊦ fs0[ri0[inodekey(ino)]].nlink = 0, adr1 = adr4, adr1 = adr3, adr1 = adr2, adr2 = ri0[inodekey(ino)], adr2 = adr4, adr2 = adr3, adr3 = ri0[inodekey(ino)], adr3 = adr4, adr4 = ri0[inodekey(ino)], ino = m1, ino = ROOT_INO, m = ino, m = 0, m1 = 0, m1 = ROOT_INO, mkdentry(str, m2) = mkdentry(str0, m1), inodekey(ino) ∈ ro, adr3 ∈ (ri0 -- dentrykey(m, str))[dentrykey(m, str0), adr2], inodekey(m) ∈ ro, adr1 ∈ fs0, adr2 ∈ fs0, adr3 ∈ fs0, adr4 ∈ fs0, adr2 ∈ ri0 -- dentrykey(m, str), adr4 ∈ (ri0 -- dentrykey(m, str))[dentrykey(m, str0), adr2][inodekey(m), adr3]

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri0[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]; (s)% ⊦ key ∈ ri ∧ ri-cons(ri, fs) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri0[inodekey(ino)]