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Proof step 13 from proof index-update-in (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: ax0, key, ri[key0, adr] abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ key ∈ ri[key0, adr] ↔ key = key0 ∨ key ∈ ri; (ws) Used terms: ax0, key, ri[key0, adr]