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Proof step 18 from proof keys-remove-leb (previous step)

@(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

User applied rule all left

And obtained the following new goals

Proof step 19:
@(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

Rule argument

pos: left 2 terms: key abort: F discard: T computed: T

Additional information

Used terms: key