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Proof step 3 from proof keys-at-other (previous step)

# kx = # ax, key ∈ ri ⊦ ri[kx, ax][key] = ri[key], key ∈ kx

User applied rule structural induction

And obtained the following new goals

Proof step 4:
#(key0 + kx0) = # ax, key ∈ ri, ∀ ax, key, ri. # kx0 = # ax ∧ key ∈ ri → ri[kx0, ax][key] = ri[key] ∨ key ∈ kx0 ⊦ ri[key0 + kx0, ax][key] = ri[key], key ∈ key0 + kx0

Rule argument

indvar: kx

Additional information

Used simplifier rules: index-update-key-base: ⊦ ri[[], []] = ri;%key-list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ # ax = 0 ↔ ax = []; (s)% ⊦ ¬ key ∈ []; (s)