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Proof step 3 from proof keys-drop-fs (previous step)

⊦ keys(fs[adr, nd], ax) = keys(fs, ax), adr ∈ ax

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ fs, adr, nd. keys(fs[adr, nd], ax0) = keys(fs, ax0) ∨ adr ∈ ax0 ⊦ keys(fs[adr, nd], adr0 + ax0) = keys(fs, adr0 + ax0), adr ∈ adr0 + ax0

Rule argument

term: ax

Additional information

Used simplifier rules: keys-base: ⊦ keys(fs, []) = [];%address-list freely generated by [], +;;