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Proof step 71 from proof ubifs-cons-reduce (previous step)

fs0[ri0[dentrykey(ino, str)]].key = dentrykey(ino, str), fs0[ri0[dentrykey(ino, str)]].dentry?, ro0.inodes?, fo0.inodes?, log-cons(log0, fs0), log-cons(log1, fs1), node-closed(fs0[ri0[dentrykey(ino, str)]], ri0, fs0, ro0), ri-cons(ri1, fs1), dentrykey(ino, str) ∈ ri0, dentrykey(ino, str) ∈ ri1, ri0[dentrykey(ino, str)] ∈ fs0, fo1 ⊆ dom fi1, ro1 ⊆ dom ri1, ro0 ⊆ dom ri0, fo0 ⊆ dom fi0, ∀ adr. ( adr ∈ log0 → (fs0[adr].inode? → fs0[adr].key.inode? ∧ fs0[adr].key.ino ≠ 0) ∧ (fs0[adr].dentry? → fs0[adr].key.dentry?) ∧ (fs0[adr].data? → fs0[adr].key.data?) ∧ (fs0[adr].trunc? → fs0[adr].key.inode?)) → adr ∈ log1 → adr ∈ fs1 ∧ (fs1[adr].inode? → fs1[adr].key.inode? ∧ fs1[adr].key.ino ≠ 0) ∧ (fs1[adr].dentry? → fs1[adr].key.dentry?) ∧ (fs1[adr].data? → fs1[adr].key.data?) ∧ (fs1[adr].trunc? → fs1[adr].key.inode?), ∀ key. key ∈ fo1 → key ∈ fi1, ∀ key. key ∈ ro1 → key ∈ ri1 ⊦ dups(log0), node-closed(fs1[ri1[dentrykey(ino, str)]], ri1, fs1, ro1), inodekey(0) ∈ ri1, inodekey(0) ∈ ri0, inodekey(ino) ∈ ri0 ∧ inodekey(fs0[ri0[dentrykey(ino, str)]].ino) ∈ ri0 ∧ ¬ inodekey(ino) ∈ ro0 ∧ ¬ inodekey(fs0[ri0[dentrykey(ino, str)]].ino) ∈ ro0 ∧ fs0[ri0[inodekey(ino)]].directory

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: node name: elim-02 seq: ⊦ nd.dentry? → (key = nd.key ∧ m = nd.ino ↔ nd = dentrynode(key, m)) subst: [nd, key, m] → [fs0[ri0[dentrykey(ino, str)]], key0, m]

Additional information

Used simplifier rules: ⊦ dentrynode(key, m).dentry?; (s)%⊦ node-closed(dentrynode(key, ino), ri, fs, ro) ↔ inodekey(key.ino) ∈ ri ∧ ¬ inodekey(key.ino) ∈ ro ∧ fs[ri[inodekey(key.ino)]].directory ∧ inodekey(ino) ∈ ri ∧ ¬ inodekey(ino) ∈ ro; (s)% ⊦ dentrykey(m, str).ino = m; (s)% ⊦ nd.dentry? → (key = nd.key ∧ m = nd.ino ↔ nd = dentrynode(key, m)); (elim) Used terms: fs0[ri0[dentrykey(ino, str)]], key0, m