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Proof step 2 from proof ri-cons-subseteq (previous step)

ri-cons(ri, fs0), fs0 ⊆ fs1 ⊦ ri-cons(ri, fs1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
key-cons(key, fs0[ri[key]]), ri-cons(ri, fs0), ri[key] ∈ fs0, key ∈ ri, ri[key] ∈ fs1, fs0 ⊆ fs1 ⊦ key-cons(key, fs1[ri[key]])

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ fs ∧ ¬ adr ∈ fs0 → ¬ fs ⊆ fs0; (s)%⊦ 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]]))); (ss)