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Proof step 19 from proof keys-remove-leb (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
key-cons(key, fs[@(n, m0, m1)]), @(n, m0, m1) ∈ ax, @(n, m0, m1) ∈ fs ⊦ fs[@(n, m0, m1)].key = key, key ∈ ri0

Rule argument

Additional information

Used simplifier rules: ⊦ ri[key, adr][key] = adr; (s)% ⊦ key ∈ ri[key, adr]; (s)