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Proof step 5 from proof ri-cons-truncnode (previous step)

fs[ri[key]] = truncnode(key0, m), key-cons(key, truncnode(key0, m)), key ∈ ri, ri[key] ∈ fs, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ key-cons(key, nd) ↔ nd.key = key ∧ (key.inode? → nd.inode? ∧ key.ino ≠ 0) ∧ (key.dentry? → nd.dentry?) ∧ (key.data? → nd.data?)
Proof step 7:
truncnode(key0, m).key = key, fs[ri[key]] = truncnode(key0, m), ri[key] ∈ fs, key ∈ ri, key.inode? → truncnode(key0, m).inode? ∧ key.ino ≠ 0, key.dentry? → truncnode(key0, m).dentry?, key.data? → truncnode(key0, m).data?, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦

Rule argument

spec: <toplevel> name: key-cons seq: ⊦ key-cons(key, nd) ↔ nd.key = key ∧ (key.inode? → nd.inode? ∧ key.ino ≠ 0) ∧ (key.dentry? → nd.dentry?) ∧ (key.data? → nd.data?) subst: [nd, key] → [truncnode(key0, m), key] flags: <all paths>

Additional information

Used terms: truncnode(key0, m), key