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Proof step 4 from proof index-update-in (previous step)

#(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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# kx0 + 1 = # ax, ∀ ax, key, ri. # kx0 = # ax → (key ∈ ri[kx0, ax] ↔ key ∈ ri ∨ key ∈ kx0) ⊦ key ∈ ri[key0 ' + kx0, ax] ↔ key ∈ ri ∨ key = key0 ∨ key ∈ kx0

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ key0 ' + kx ↔ key = key0 ∨ key ∈ kx; (s)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ m +1 = m + 1; (s)% ⊦ #(key ' + kx) = (# kx)+1; (s)% ⊦ key + kx = key ' + kx; (s)