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Proof step 143 from proof inv-link (previous step)

fs0[ri0[inodekey(ino1)]].nlink = 0, fs0[ri0[inodekey(fs0[ri0[dentrykey(ino, str)]].ino)]] = inodenode(key1, md2, false, m4, m5, m6), fs0[ri0[dentrykey(ino0, str)]] = dentrynode(key, fs0[ri0[dentrykey(ino, str)]].ino), fs0[ri0[inodekey(m3)]] = inodenode(key0, md1, true, m, m0, m1), fs0[ri0[inodekey(ino)]].directory, fs0[ri0[inodekey(ino0)]].directory, 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), ri0[inodekey(m3)] ∈ fs0, ri0[inodekey(fs0[ri0[dentrykey(ino, str)]].ino)] ∈ fs0, inodekey(ino) ∈ ri0, inodekey(ino0) ∈ ri0, dentrykey(ino, str) ∈ ri0, ri0[dentrykey(ino0, str)] ∈ fs0, inodekey(fs0[ri0[dentrykey(ino, str)]].ino) ∈ ri0, inodekey(m3) ∈ ri0, dentrykey(ino0, str) ∈ ri0, inodekey(ino1) ∈ ri0, inodekey(ino1) ∈ ro, fs0[ri0[inodekey(ino)]].nlink ≠ 0 ∨ ino = ROOT_INO, m ≠ 0 ∨ m3 = ROOT_INO, fs0[ri0[inodekey(ino0)]].nlink ≠ 0 ∨ ino0 = ROOT_INO ⊦ # ro = 0, fs0[ri0[dentrykey(ino, str)]].ino = 0, (adr1 = ri0[inodekey(ino1)] ⊃ inodenode(inodekey(m3), md0, true, m, m0, m1 + 1);fs0[ri0[inodekey(ino1)]]).nlink = 0, adr1 = adr2, adr1 = adr3, adr2 = ri0[inodekey(ino1)], adr2 = adr3, adr3 = ri0[inodekey(ino1)], ino1 = ROOT_INO, m3 = 0, ro = ∅, adr3 ∈ fs0, adr2 ∈ fs0, adr1 ∈ fs0, inodekey(ino) ∈ ro, inodekey(fs0[ri0[dentrykey(ino, str)]].ino) ∈ ro, dentrykey(m3, str0) ∈ ri0, inodekey(m3) ∈ ro, inodekey(ino0) ∈ ro

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri0[inodekey(ino1)]

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]; (s)% ⊦ inodenode(key, md, boolvar, m, m0, m1).nlink = m; (s)% ⊦ key ∈ ri ∧ ri-cons(ri, fs) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri0[inodekey(ino1)]