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Proof step 226 from proof inv-rename-new-reparent (previous step)

n0 + (boolvar2 ⊃ 1;0) = fs[ri[inodekey(m)]].nsubdirs, m1 + 1 = fs[ri[inodekey(m)]].size, fs[ri[inodekey(m3)]] = inodenode(key0, md1, boolvar0, m5, m9, m10), fs[ri[dentrykey(m, str)]] = dentrynode(key, m3), adr5 = adr1, adr6 = adr2, m2 = fs[ri[inodekey(m)]].nlink, m6 = fs[ri[inodekey(m0)]].nlink, m7 = fs[ri[inodekey(m0)]].nsubdirs, m8 = fs[ri[inodekey(m0)]].size, nd = inodenode(inodekey(m0), md0, true, m6, m7 + (boolvar2 ⊃ 1;0), m8 + 1), nd0 = inodenode(inodekey(m), md, true, m2, n0, m1), nd1 = dentrynode(dentrykey(m0, str0), mkdentry(str, m3).ino), nd2 = dentrynode(dentrykey(m, str), 0), mkdentry(str, m3).dentry?, fs[ri[inodekey(m0)]].directory, fs[ri[inodekey(m)]].directory, ro.inodes?, fo.inodes?, file-node-cons(fs, ri), log-cons(log, fs), orphans-cons(fs, ri, ro), replay-cons-strong(fs, ri, ro, fi, fo, log), ri-inj(ri), ri-inj(fi), ubifs-cons(fs, ri, ro, fi, fo, log), ri[inodekey(m3)] ∈ fs, ri[dentrykey(m, str)] ∈ fs, inodekey(m) ∈ ri, inodekey(m0) ∈ ri, inodekey(m3) ∈ ri, dentrykey(m, str) ∈ ri, ∀ adr1, adr2, adr3, adr4. ¬(new(adr1 ' + adr2 ' + (adr3 ' + adr4), fs) ∧ ¬ dups(adr1 ' + adr2 ' + adr3 ' + adr4 ')), ¬ boolvar0 ∧ ¬ boolvar2 ∨ boolvar0 ∧ boolvar2 ∧ m4 = m10, fs[ri[inodekey(m0)]].nlink ≠ 0 ∨ m0 = ROOT_INO, fs[ri[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO ⊦ ⟪skip⟫ (ubifs-cons(fs, ri, ro, fi, fo, log) ∧ replay-cons-strong(fs, ri, ro, fi, fo, log) ∧ file-node-cons(fs, ri) ∧ orphans-cons(fs, ri, ro)), m0 = m, inodekey(m0) ∈ ro, dentrykey(m0, str0) ∈ ri, inodekey(m) ∈ ro

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkdentry(str, m).dentry?; (s)% ⊦ ax + adr = ax + adr '; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ new(adr ' + log, fs) ↔ ¬ adr ∈ fs ∧ new(log, fs); (s)% ⊦ new(adr ', fs) ↔ ¬ adr ∈ fs; (s)% ⊦ dups(adr ' + ax) ↔ adr ∈ ax ∨ dups(ax); (s)% ⊦ adr ∈ adr0 ' + ax ↔ adr = adr0 ∨ adr ∈ ax; (s)% ⊦ adr ∈ adr0 ' ↔ adr = adr0; (s)% ⊦ dups(adr ' + adr0 ') ↔ adr = adr0; (s)