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Proof step 2 from proof ri-inj-variant1-01 (previous step)

ri-inj(ri) ⊦ ri-inj(ri0) ↔ (∀ key0, key1. (key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1) → key0 ∈ ri0 ∧ key1 ∈ ri0 ∧ ri0[key0] = ri0[key1] → key0 = key1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ (∀ key0, key1. key0 ∈ ri0 ∧ key1 ∈ ri0 ∧ ri0[key0] = ri0[key1] → key0 = key1) ↔ (∀ key, key0. ¬ key ∈ ri ∨ ¬ key0 ∈ ri ∨ ri[key] ≠ ri[key0] → key ∈ ri0 ∧ key0 ∈ ri0 ∧ ri0[key] = ri0[key0] → key = key0)

Rule argument

Additional information

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