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Proof step 697 from proof inv-rename-overwrite-keep-parent (previous step)

fs[ri[inodekey(m8)]] = inodenode(key, md0, boolvar0, m0, m9, m10), fs[ri[dentrykey(m, str0)]] = dentrynode(key0, m8), adr5 = adr1, adr6 = adr2, m1 = fs[ri[inodekey(m)]].nlink, m2 = fs[ri[inodekey(m)]].nsubdirs, m4 = fs[ri[inodekey(m)]].size, nd = inodenode(inodekey(m8), md1, boolvar2, m0, m9, m6), nd0 = inodenode(inodekey(m), md, true, m1, m2, m4), nd1 = dentrynode(dentrykey(m, str0), mkdentry(str0, m8).ino), nd2 = dentrynode(dentrykey(m, str0), 0), mkdentry(str0, m8).dentry?, fs[ri[inodekey(m)]].directory, fo.inodes?, ro.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(fi), ri-inj(ri), ubifs-cons(fs, ri, ro, fi, fo, log), dentrykey(m, str0) ∈ ri, inodekey(m8) ∈ ri, inodekey(m) ∈ ri, ri[dentrykey(m, str0)] ∈ fs, ri[inodekey(m8)] ∈ fs, 1 ≤ fs[ri[inodekey(m)]].size, boolvar3 → 1 ≤ fs[ri[inodekey(m)]].nsubdirs, m0 = 1 → links(m8, fs, ri) = ∅ ++ dentrykey(m, str0), boolvar0 → dentrykeys(inodekey(m8), ri) = ∅, ∀ adr1, adr2, adr3, adr4. ¬(new(adr1 ' + adr2 ' + (adr3 ' + adr4), fs) ∧ ¬ dups(adr1 ' + adr2 ' + adr3 ' + adr4 ')), fs[ri[inodekey(m)]].nlink ≠ 0 ∨ m = ROOT_INO, ¬ boolvar0 ∧ ¬ boolvar3 ∨ boolvar0 ∧ boolvar3 ∧ m3 = m10, ¬ boolvar0 ∧ ¬ boolvar2 ∨ boolvar0 ∧ boolvar2 ∧ m6 = m10 ⊦ ⟪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)), fs[ri[inodekey(m)]].size = 0, m0 = 0, m8 = ROOT_INO, 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)