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Proof step 16 from proof keys-remove-leb (previous step)

ri-cons(ri0[key, @(n, m0, m1)], fs), @(n, m0, m1) ∈ ax ⊦ fs[@(n, m0, m1)].key = key, key ∈ ri0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ ri-cons(ri, fs) ↔ (∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]))
Proof step 18:
@(n, m0, m1) ∈ ax, ∀ key0. key0 ∈ ri0[key, @(n, m0, m1)] → ri0[key, @(n, m0, m1)][key0] ∈ fs ∧ key-cons(key0, fs[ri0[key, @(n, m0, m1)][key0]]) ⊦ fs[@(n, m0, m1)].key = key, key ∈ ri0

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] → [fs, ri0[key, @(n, m0, m1)]] flags: <all paths>

Additional information

Used terms: fs, ri0[key, @(n, m0, m1)]