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Proof step 8 from proof size-in (previous step)

st = (st -- a)[a, st[a]] ⊦ a ∈ st, # st[a, d] = # st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
st = (st -- a)[a, st[a]] ⊦ a ∈ st

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ st → # st[a, d] = # st + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2