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Proof step 9 from proof ri-inj-variant1-01 (
previous step
)
ri0[key0] = ri0[key], ri[key0] = ri[key], key0 ∈ ri0, key ∈ ri0, key0 ∈ ri, key ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ key0 = key
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 7 terms: key0, key
Additional information
Used terms: key0, key