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

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

Heuristic quantifier applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 4 terms: key0, key0

Additional information

Used terms: key0, key0