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Proof step 19 from proof range-del-in (
previous step
)
key0 ∈ ri, key ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ ri[key0] = ri[key], ∃ key1. key1 ∈ ri ∧ ri[key1] = ri[key0]
Heuristic quantifier closing applied rule exists right
And closed the goal!
Rule argument
pos: right 2 terms: key0
Additional information
Used terms: key0