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

fs[ri[key]] = truncnode(key0, m), 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]]) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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