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Proof step 2 from proof in-remove-refsize0-refsize-cons (previous step)

inv(st), refsize-cons(st, as), adr ∈ as, adr ∈ st ⊦ adr ∈ remove-refsize0(st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
inv(st), adr ∈ as, adr ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, adr ∈ remove-refsize0(st)

Rule argument

Additional information

Used simplifier rules: refsize-cons-def: ⊦ refsize-cons(st, as) ↔ (∀ n. n ∈ st → st[n].ref-size = (as | n).size);%size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[address-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[address-store]