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Proof step 130 from proof inv-rename-new-keep-parent (previous step)

fs0[ri0[dentrykey(m, str)]] = dentrynode(key0, m0), fs0[ri0[inodekey(m)]] = inodenode(key1, md0, true, m4, m5, m6), replay-log(log0, fs0, fi ⊖ fo) = ri0 ⊖ ro, ro.inodes?, fo.inodes?, file-node-cons(fs0, ri0), log-cons(log0, fs0), orphans-cons(fs0, ri0, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), replay-safe(log0, fs0, fi ⊖ fo), ri-inj(ri0), ri-inj(fi), ubifs-cons(fs0, ri0, ro, fi, fo, log0), ri0[inodekey(m)] ∈ fs0, inodekey(m) ∈ ri0, ri0[dentrykey(m, str)] ∈ fs0, inodekey(m0) ∈ ri0, dentrykey(m, str) ∈ ri0, 1 ≤ m6, boolvar2 → 1 ≤ m5, m4 ≠ 0 ∨ m = ROOT_INO, ¬ fs0[ri0[inodekey(m0)]].directory ∧ ¬ boolvar2 ∨ fs0[ri0[inodekey(m0)]].directory ∧ boolvar2 ∧ m3 = fs0[ri0[inodekey(m0)]].size, key = dentrykey(m, str0) ∨ key ∈ ri0 ⊖ ro ⊦ adr1 = adr2, adr1 = adr3, adr2 = adr3, key = inodekey(m), key = dentrykey(m, str), m = 0, m0 = 0, m6 = 0, adr2 ∈ ri0 -- dentrykey(m, str), adr3 ∈ fs0, adr2 ∈ fs0, adr1 ∈ fs0, dentrykey(m, str0) ∈ ri0, inodekey(m) ∈ ro, adr3 ∈ (ri0 -- dentrykey(m, str))[dentrykey(m, str0), adr2], adr3 ∈ log0, adr2 ∈ log0, adr1 ∈ log0, (key = dentrykey(m, str0) ∨ key ∈ ri0) ∧ ¬ key ∈ ro ∧ (key.data? → ¬ inodekey(key.ino) ∈ ro)

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: key = dentrykey(m, str0)

Additional information

Used simplifier rules: ⊦ key ∈ ri ⊖ ks ↔ key ∈ ri ∧ ¬ key ∈ ks ∧ (key.data? → ¬ inodekey(key.ino) ∈ ks); (ws)% ⊦ inodekey(m) ≠ dentrykey(m0, str); (s)% ⊦ ks.inodes? → ¬ dentrykey(ino, str) ∈ ks; (s)% ⊦ ¬ dentrykey(m, str).data?; (s) Used formulas: key = dentrykey(m, str0)