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Proof step 6 from proof key-node-cons (previous step)

key = inodekey(ino), key-cons(key, nd) ⊦ (nd.inode? → key.inode? ∧ key.ino ≠ 0) ∧ (nd.dentry? → key.dentry?) ∧ (nd.data? → key.data?) ∧ (nd.trunc? → key.inode?)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
nd.key = inodekey(ino), nd.inode? ⊦ ino = 0, ¬ nd.dentry? ∧ ¬ nd.data?

Rule argument

Additional information

Used simplifier rules: inodekey-cons: ⊦ key-cons(inodekey(ino), nd) ↔ nd.inode? ∧ nd.key = inodekey(ino) ∧ ino ≠ 0;% ⊦ ¬ inodekey(m).data?; (s)% ⊦ ¬ inodekey(m).dentry?; (s)% ⊦ inodekey(m).ino = m; (s)% ⊦ inodekey(m).inode?; (s)