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Proof step 5 from proof ri-inj-variant1-01 (previous step)

ri0[key0] = ri0[key], key0 ∈ ri0, key ∈ ri0, ∀ key0, key1. key0 ∈ ri0 ∧ key1 ∈ ri0 ∧ ri0[key0] = ri0[key1] → key0 = key1, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1, ¬ key0 ∈ ri ∨ ¬ key ∈ ri ∨ ri[key0] ≠ ri[key] ⊦ key0 = key

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: key0, key

Additional information

Used terms: key0, key