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Proof step 2 from proof remove-refsize0-subset-refed (previous step)

refsize-cons(st, as), valid-adrs(as), as ⊆ dom nodes(st) ⊦ nodes(st) | as ⊆ nodes(remove-refsize0(st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
as ⊆ dom nodes(st), ∀ adr. adr ∈ as → adr.size ≠ 0, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ nodes(st) | as ⊆ nodes(remove-refsize0(st))

Rule argument

Additional information

Used simplifier rules: valid-adrs: ⊦ valid-adrs(as) ↔ (∀ adr. adr ∈ as → adr.size ≠ 0);%refsize-cons-def: ⊦ refsize-cons(st, as) ↔ (∀ n. n ∈ st → st[n].ref-size = (as | n).size);