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Proof step 1 from proof in-ri-dentry-inode (conclusion)

⊦ dentrykey(ino, str) ∈ ri[inodekey(ino), adr] ↔ dentrykey(ino, str) ∈ ri

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ inodekey(m) ≠ dentrykey(m0, str); (s)% ⊦ key ∈ ri[key0, adr] ↔ key = key0 ∨ key ∈ ri; (ws)