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Proof step 36 from proof replay-one-update (previous step)

fs[ri[dentrykey(ino, str)]].key = dentrykey(ino, str), fs[ri[dentrykey(ino, str)]].dentry?, ro.inodes?, orphans-cons(fs, ri, ro), ri-closed(ri, fs, ro), dentrykey(ino, str) ∈ ri, ri[dentrykey(ino, str)] ∈ fs, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ replay-one(adr, fs[ri[dentrykey(ino, str)]], ri ⊖ ro) = ri[dentrykey(ino, str), adr] ⊖ ro

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 37:
fs[ri[dentrykey(ino, str)]] = dentrynode(key0, m), key0 = dentrykey(ino, str), fs[ri[dentrykey(ino, str)]].dentry?, ro.inodes?, orphans-cons(fs, ri, ro), ri-closed(ri, fs, ro), ri[dentrykey(ino, str)] ∈ fs, dentrykey(ino, str) ∈ ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ replay-one(adr, fs[ri[dentrykey(ino, str)]], ri ⊖ ro) = ri[dentrykey(ino, str), adr] ⊖ ro

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