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Proof step 10 from proof key-node-cons (previous step)

key = dentrykey(ino, str), 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 11:
nd.key = dentrykey(ino, str), nd.dentry? ⊦ ¬ nd.inode? ∧ ¬ nd.data? ∧ ¬ nd.trunc?

Rule argument

Additional information

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