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Proof step 15 from proof keys-remove-leb (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
ri-cons(ri0[key, @(n, m0, m1)], fs), @(n, m0, m1) ∈ ax ⊦ fs[@(n, m0, m1)].key = key, key ∈ ri0

Proof step 15 from proof keys-remove-leb (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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