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Proof step 369 from proof inv-unlink (previous step)

m3 + 1 = fs[ri[inodekey(fs[ri[dentrykey(m1, str)]].ino)]].nlink, fs[ri[inodekey(m1)]].directory, fs[ri[inodekey(m1)]].inode?, fs[ri[inodekey(fs[ri[dentrykey(m1, str)]].ino)]].inode?, 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-cons(ri, fs), ri-inj(fi), ri-inj(ri), ubifs-cons(fs, ri, ro, fi, fo, log), inodekey(m1) ∈ ri, ri[inodekey(m1)] ∈ fs, dentrykey(m1, str) ∈ ri, ri[inodekey(fs[ri[dentrykey(m1, str)]].ino)] ∈ fs, inodekey(fs[ri[dentrykey(m1, str)]].ino) ∈ 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 = 0, fs[ri[dentrykey(m1, str)]].ino = ROOT_INO, fs[ri[inodekey(m1)]].size = 0, m1 = 0, fs[ri[inodekey(fs[ri[dentrykey(m1, str)]].ino)]].directory, inodekey(fs[ri[dentrykey(m1, str)]].ino) ∈ ro, inodekey(m1) ∈ ro, ∃ err0. ⌊err0⌋

Proof step 369 from proof inv-unlink (previous step)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

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