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Proof step 10 from proof ubifs-parent-in (previous step)

fs[ri[dentrykey(ino, str)]].key = dentrykey(ino, str), fs[ri[dentrykey(ino, str)]].dentry?, fo.inodes?, ro.inodes?, node-closed(fs[ri[dentrykey(ino, str)]], ri, fs, ro), dentrykey(ino, str) ∈ ri, ri[dentrykey(ino, str)] ∈ fs, fo ⊆ dom fi, ro ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ node-closed(fs[ri[key]], ri, fs, ro), ∀ adr. adr ∈ log → adr ∈ fs ∧ node-cons(fs[adr].key, fs[adr]) ⊦ dups(log), inodekey(0) ∈ ri, inodekey(ino) ∈ ri

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] → [fs[ri[dentrykey(ino, str)]], key0, m]

Additional information

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