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Proof step 299 from proof inv-unlink (previous step)

fs0[ri0[inodekey(m1)]] = inodenode(key0, md1, true, m4, m5, m + 1), fs0[ri0[dentrykey(m1, str)]] = dentrynode(key, m0), fs0[ri0[inodekey(m0)]] = inodenode(key1, md2, false, m3 + 1, m7, m8), replay-log(log0, fs0, fi ⊖ fo) = ri0 ⊖ ro, 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), replay-safe(log0, fs0, fi ⊖ fo), ri-cons(ri0, fs0), ri-inj(fi), ri-inj(ri0), ubifs-cons(fs0, ri0, ro, fi, fo, log0), key2 ∈ ri0, dentrykey(m1, str) ∈ ri0, ri0[dentrykey(m1, str)] ∈ fs0, inodekey(m0) ∈ ri0, ri0[inodekey(m0)] ∈ fs0, inodekey(m1) ∈ ri0, ri0[inodekey(m1)] ∈ fs0, m4 ≠ 0 ∨ m1 = ROOT_INO, key2 = inodekey(m0) ∨ key2 ∈ ri0[inodekey(m1), adr1] ⊖ ro ⊦ adr1 = adr3, adr1 = adr2, adr2 = adr3, key2 = dentrykey(m1, str), m0 = 0, m0 = ROOT_INO, m1 = 0, m3 = 0, adr1 ∈ log0, adr2 ∈ log0, adr3 ∈ log0, inodekey(m1) ∈ ro, inodekey(m0) ∈ ro, adr1 ∈ fs0, adr2 ∈ fs0, adr3 ∈ fs0, ¬ key2 ∈ ro ∧ (key2.data? → ¬ inodekey(key2.ino) ∈ ro)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 17

Additional information

Used simplifier rules: in-ri: ⊦ key0 ∈ ri[key1, adr] ↔ key0 = key1 ∨ key0 ∈ ri;%elim-data: ⊦ key.data? ↔ (∃ ino, n. key = datakey(ino, n));%in-ri: ⊦ key0 ∈ ri[key1, adr] ↔ key0 = key1 ∨ key0 ∈ ri;%ne-01: ⊦ key ∈ ks → ks ≠ ∅; (forward) from specification set-basic[node-index]% ⊦ datakey(m, m0) ≠ dentrykey(m1, str); (s)% ⊦ datakey(m, m0).ino = m; (s)% ⊦ inodekey(m) ≠ datakey(m0, m1); (s)%⊦ key ∈ ri ⊖ ks ↔ key ∈ ri ∧ ¬ key ∈ ks ∧ (key.data? → ¬ inodekey(key.ino) ∈ ks); (ws)% ⊦ ks.inodes? → ¬ datakey(ino, n) ∈ ks; (s)% ⊦ datakey(m, m0).data?; (s)