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Proof step 7 from proof keys-at-other (previous step)

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

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