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Proof step 221 from proof inv-truncate (previous step)

# buf-opt.buf = VFS_PAGE_SIZE, fs0[ri0[inodekey(m)]].nlink = 0 ∧ m ≠ ROOT_INO ↔ inodekey(m) ∈ ro, VFS_PAGE_SIZE = n0 + m3 % VFS_PAGE_SIZE, fs0[ri0[inodekey(m)]].inode?, ro.inodes?, fo.inodes?, buf-opt.some?, file-node-cons(fs0, ri0), log-cons(log0, fs0), orphans-cons(fs0, ri0, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), ri-inj(ri0), ri-inj(fi), ubifs-cons(fs0, ri0, ro, fi, fo, log0), ri0[inodekey(m)] ∈ fs0, inodekey(m) ∈ ri0, inodekey(ino) ∈ ri0, m3 ≤ n1, ¬ fs0[ri0[inodekey(ino)]].directory → fs0[ri0[inodekey(ino)]].nsubdirs = 0 ⊦ m3 % VFS_PAGE_SIZE = 0, (adr1 = ri0[inodekey(ino)] ⊃ truncnode(inodekey(m), m3);fs0[ri0[inodekey(ino)]]).nsubdirs = 0, adr1 = adr3, adr1 = adr2, adr2 = ri0[inodekey(ino)], adr2 = adr3, adr3 = ri0[inodekey(ino)], m = ino, fs0[ri0[inodekey(m)]].directory, (adr1 = ri0[inodekey(ino)] ⊃ truncnode(inodekey(m), m3);fs0[ri0[inodekey(ino)]]).directory, adr1 ∈ fs0, adr2 ∈ fs0, adr3 ∈ fs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri0[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ key ∈ ri ∧ ubifs-cons(fs, ri, ro, fi, fo, log) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri0[inodekey(ino)]