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Proof step 10 from proof subset-size (previous step)

# st0 < # st1, a ∈ st1, st1 -- a ⊆ st0 -- a ∧ a × d ∈ st0, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a, d] = ∅, st1 = ∅, st1[a] = d

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a ∈ st0, st1 -- a ⊆ st0 -- a, a ∈ st1, # st0 < # st1, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a] = st0[a], st1 = ∅, st1[a, st0[a]] = ∅

Rule argument

Additional information

Used simplifier rules: in-pair: ⊦ a × d ∈ st ↔ ¬ ¬(a ∈ st ∧ st[a] = d);