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Proof step 2 from proof ri-inj-del (previous step)

ri-inj(ri) ⊦ ri-inj(ri -- key)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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

Rule argument

Additional information

Used simplifier rules: ri-inj: ⊦ ri-inj(ri) ↔ (∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1);% ⊦ key ≠ key0 → (ri -- key0)[key] = ri[key]; (s)% ⊦ key ∈ ri -- key0 ↔ key ≠ key0 ∧ key ∈ ri; (s)