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Proof step 14 from proof keys-remove-leb (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 15:
@(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

Rule argument

pos: left 2 terms: @(n, m0, m1) abort: F discard: T computed: T

Additional information

Used terms: @(n, m0, m1)