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Proof step 3 from proof datakeys-set-add (previous step)

⊦ ∀ key0. key0 ∈ datakeys(ks ++ key, ri) ↔ key0 ∈ datakeys(ks, ri) ∪ datakeys(key, ri)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ key0 ∈ ri ∧ key0.data? ∧ (inodekey(key0.ino) = key ∨ inodekey(key0.ino) ∈ ks) ↔ key0 ∈ datakeys(ks, ri) ∪ datakeys(key, ri)

Rule argument

Additional information

Used simplifier rules: datakeys-set: ⊦ key ∈ datakeys(ks, ri) ↔ key ∈ ri ∧ key.data? ∧ inodekey(key.ino) ∈ ks;% ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)