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Proof step 8 from proof range-del-in (previous step)

ri-inj(ri), key ∈ ri ⊦ (∃ key0. key0 ∈ ri -- key ∧ (ri -- key)[key0] = adr) ↔ adr ≠ ri[key] ∧ adr ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
ri-inj(ri), key ∈ ri ⊦ (∃ key0. key0 ≠ key ∧ key0 ∈ ri ∧ ri[key0] = adr) ↔ adr ≠ ri[key] ∧ adr ∈ ri

Rule argument

Additional information

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