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

refsize-cons(st, as) ⊦ refsize-cons(remove-refsize0(st), as)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ st[n].ref-size = 0, st[n].ref-size = (as | n).size

Rule argument

Additional information

Used simplifier rules: refsize-cons-def: ⊦ refsize-cons(st, as) ↔ (∀ n. n ∈ st → st[n].ref-size = (as | n).size);% ⊦ n ∈ st ∧ st[n].ref-size ≠ 0 → remove-refsize0(st)[n] = st[n]; (s)% ⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0; (ws)