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Proof step 19 from proof index-update-in (previous step)

# kx0 = # ax0, key ∈ kx0, ∀ ax, key, ri. # ax0 = # ax → (key ∈ ri[kx0, ax] ↔ key ∈ ri ∨ key ∈ kx0) ⊦ key = key0, key ∈ ri, key ∈ ri[key0, adr][kx0, ax0]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: ax0, key, ri[key0, adr]

Additional information

Used simplifier rules: ⊦ key ≠ key0 → (key ∈ ri[key0, adr] ↔ key ∈ ri); (s) Used terms: ax0, key, ri[key0, adr]