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Proof step 3 from proof ri-cons-subseteq (previous step)

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]])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]))
Proof step 5:
key-cons(key, fs0[ri[key]]), ri[key] ∈ fs1, key ∈ ri, ri[key] ∈ fs0, fs0 ⊆ fs1, ∀ key. key ∈ ri → ri[key] ∈ fs0 ∧ key-cons(key, fs0[ri[key]]) ⊦ key-cons(key, fs1[ri[key]])

Rule argument

spec: aubifs-inv name: ri-cons seq: ⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]])) subst: [fs, ri] → [fs0, ri] flags: <all paths>

Additional information

Used terms: fs0, ri