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Proof step 3 from proof index-update-in (previous step)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

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

Rule argument

term: 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)