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Proof step 3 from proof ri-inj-del (previous step)

ri[key0] = ri[key1], key0 ∈ ri, key1 ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ key0 = key1, key0 = key, key1 = key

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
ri[key0] = ri[key1], key1 ∈ ri, key0 ∈ ri, key1 ∈ ri ∧ key1 ∈ ri ∧ ri[key1] = ri[key1] → key1 = key1, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ key0 = key, key0 = key1, key1 = key

Rule argument

pos: left 4 terms: key1, key1

Additional information

Used terms: key1, key1