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Proof step 349 from proof inv-unlink (previous step)

(adr1 = ri0[inodekey(ino)] ⊃ inodenode(inodekey(m1), md0, true, m0, m4, m5 - 1);fs0[ri0[inodekey(ino)]]).nlink = 0, fs0[ri0[inodekey(m1)]] = inodenode(key0, md1, true, m0, m4, m5), fs0[ri0[dentrykey(m1, str)]] = dentrynode(key, m), fs0[ri0[inodekey(m)]] = inodenode(key1, md2, false, m3 + 1, m7, m8), 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, dentrykey(m1, str) ∈ ri0, ri0[dentrykey(m1, str)] ∈ fs0, inodekey(m) ∈ ri0, ri0[inodekey(m)] ∈ fs0, inodekey(m1) ∈ ri0, ri0[inodekey(m1)] ∈ fs0, 1 ≤ m5, m0 ≠ 0 ∨ m1 = ROOT_INO ⊦ fs0[ri0[inodekey(ino)]].nlink = 0, adr1 = adr3, adr1 = adr2, adr2 = ri0[inodekey(ino)], adr2 = adr3, adr3 = ri0[inodekey(ino)], ino = ROOT_INO, m = ino, m = 0, m = ROOT_INO, m1 = ino, m1 = 0, m3 = 0, m5 = 0, inodekey(ino) ∈ ro, inodekey(m1) ∈ ro, inodekey(m) ∈ ro, 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: ⊦ inodenode(key, md, boolvar, m, m0, m1).nlink = m; (s)% ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]; (s)% ⊦ key ∈ ri ∧ ri-cons(ri, fs) → ri[key] ∈ fs; (ss) Used formulas: adr1 = ri0[inodekey(ino)]