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Proof step 4 from proof ri-cons-inj (previous step)

ri[key] = ri[key0], key0 ∈ ri, key ∈ ri, key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key = key0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
ri[key] = ri[key0], key-cons(key, fs[ri[key0]]), key ∈ ri, key0 ∈ ri, ri[key0] ∈ fs, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ key = key0

Proof step 4 from proof ri-cons-inj (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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