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Proof step 2 from proof refsize-cons-dealloc (previous step)

refsize-cons(st, as) ⊦ refsize-cons(st -- n, as)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ st[n0].ref-size = (as | n0).size, n0 = n

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 ≠ n0 → (st -- n0)[n] = st[n]; (s)% ⊦ n ∈ st -- n0 ↔ n ≠ n0 ∧ n ∈ st; (s)