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Proof step 5 from proof notin-dels (previous step)

∃ key. key ∈ ri ⊖ ks ∧ (ri ⊖ ks)[key] = adr ⊦ adr ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
key.data?, key ∈ ri ⊦ key ∈ ks, (ri ⊖ ks)[key] ∈ ri, inodekey(key.ino) ∈ ks

Rule argument

Additional information

Used simplifier rules: in-dels-ro: ⊦ key ∈ ri ⊖ ks ↔ key ∈ ri ∧ ¬ key ∈ ks ∧ (key.data? → ¬ inodekey(key.ino) ∈ ks);%at-dels-ro-nodata: ⊦ key ∈ ri ∧ ¬ key ∈ ks ∧ ¬ key.data? → (ri ⊖ ks)[key] = ri[key];%key ∈ ri ⊦ ri[key] ∈ ri; (s)