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Proof step 10 from proof keys-remove-leb (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 11:
ax.toSet = (range ri0 | n) ++ @(n, m0, m1), ri-cons(ri0[key, @(n, m0, m1)], fs) ⊦ key ∈ ri0, @(n, m0, m1) ∈ ax ∧ fs[@(n, m0, m1)].key = key

Rule argument

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

Additional information

Used terms: @(n, m0, m1)