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Proof step 680 from proof inv-truncate (previous step)

m = 0 ∧ m0 ≠ ROOT_INO ↔ inodekey(m0) ∈ ro, fs0[ri0[inodekey(ino)]].nlink = 0 ↔ inodekey(ino) ∈ ro, fs0[ri0[inodekey(m0)]] = inodenode(key, md, false, m, m1, m2), fo.inodes?, ro.inodes?, file-node-cons(fs0, ri0), log-cons(log0, fs0), orphans-cons(fs0, ri0, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), ri-cons(ri0, fs0), ri-inj(fi), ri-inj(ri0), ubifs-cons(fs0, ri0, ro, fi, fo, log0), inodekey(ino) ∈ ri0, ri0[inodekey(m0)] ∈ fs0, inodekey(m0) ∈ ri0, m4 ≤ n0 ⊦ m4 % VFS_PAGE_SIZE = 0, (adr1 = ri0[inodekey(ino)] ⊃ truncnode(inodekey(m0), m4);fs0[ri0[inodekey(ino)]]).nlink = 0 ↔ inodekey(ino) ∈ ro, adr1 = adr2, adr2 = ri0[inodekey(ino)], ino = ROOT_INO, m0 = ino, m0 = 0, buf-opt.some?, datakey(m0, m4 / VFS_PAGE_SIZE) ∈ ri0, adr1 ∈ fs0, adr2 ∈ fs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr1 = ri0[inodekey(ino)]

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]; (s)% ⊦ key ∈ ri ∧ ri-cons(ri, fs) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri0[inodekey(ino)]