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Proof step 371 from proof inv-unlink (previous step)

m3 + 1 = fs[ri[inodekey(fs[ri[dentrykey(m1, str)]].ino)]].nlink, adr4 = adr1, m = fs[ri[dentrykey(m1, str)]].ino, nd = inodenode(inodekey(m), md, false, m3, fs[ri[inodekey(m)]].nsubdirs, m2), nd0 = dentrynode(dentrykey(m1, str), 0), nd1 = inodenode(inodekey(m1), md0, true, fs[ri[inodekey(m1)]].nlink, fs[ri[inodekey(m1)]].nsubdirs, fs[ri[inodekey(m1)]].size - 1), fs[ri[inodekey(m1)]].directory, fs[ri[inodekey(m)]].inode?, fs[ri[inodekey(m1)]].inode?, 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), inodekey(m) ∈ ri, ri[inodekey(m)] ∈ fs, dentrykey(m1, str) ∈ ri, inodekey(fs[ri[dentrykey(m1, str)]].ino) ∈ ri, ri[inodekey(m1)] ∈ fs, inodekey(m1) ∈ ri, 1 ≤ fs[ri[inodekey(m1)]].size, m3 = 0 → links(fs[ri[dentrykey(m1, str)]].ino, fs, ri) = ∅ ++ dentrykey(m1, str), fs[ri[inodekey(m1)]].nlink ≠ 0 ∨ m1 = ROOT_INO ⊦ fs[ri[dentrykey(m1, str)]].ino = ROOT_INO, fs[ri[inodekey(m1)]].size = 0, fs[ri[inodekey(fs[ri[dentrykey(m1, str)]].ino)]].directory, inodekey(fs[ri[dentrykey(m1, str)]].ino) ∈ ro, inodekey(m1) ∈ ro, ri-cons(ri, fs) ∧ node-cons(nd1.key, nd1) ∧ node-cons(nd0.key, nd0) ∧ node-cons(nd.key, nd)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ROOT_INO ≠ 0; (s)% ⊦ ubifs-cons(fs, ri, ro, fi, fo, log) → ¬ inodekey(0) ∈ ri; (s)% ⊦ dentrykey(m, str).dentry?; (s)% ⊦ node-cons(key0, dentrynode(key1, ino)) ↔ key0.dentry?; (s)% ⊦ dentrynode(key, m).key = key; (s)% ⊦ inodekey(m).ino = m; (s)% ⊦ inodekey(m).inode?; (s)%⊦ node-cons(key0, inodenode(key1, md, boolvar, n, n0, m)) ↔ key0.inode? ∧ key0.ino ≠ 0; (s)% ⊦ inodenode(key, md, boolvar, m, m0, m1).key = key; (s)% ⊦ ubifs-cons(fs, ri, ro, fi, fo, log) → ri-cons(ri, fs); (ss)