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

ri-cons(ri, fs) ⊦ ri-cons(ri0, fs0) ↔ (∀ key. (key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]])) → key ∈ ri0 → ri0[key] ∈ fs0 ∧ key-cons(key, fs0[ri0[key]]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
key ∈ ri0, key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri0 → ri0[key] ∈ fs0 ∧ key-cons(key, fs0[ri0[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri0[key] ∈ fs0 ∧ key-cons(key, fs0[ri0[key]])

Rule argument

Additional information

Used simplifier rules: ri-cons: ⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]));%ri-cons-in: ⊦ key ∈ ri ∧ ri-cons(ri, fs) → ri[key] ∈ fs;%ri-key-cons: ⊦ key ∈ ri ∧ ri-cons(ri, fs) → key-cons(key, fs[ri[key]]);