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Proof step 21 from proof subset-size (previous step)

st0[a] = d, a ∈ st0, # st0 ≤ # st1, st1 -- a ⊆ st0 -- a, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, d] = ∅, a ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
a ∈ st0, # st0 ≤ # st1, st1 ⊆ st0 -- a, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, st0[a]] = ∅, a ∈ st1

Rule argument

Additional information

Used simplifier rules: notin-del-notin: ⊦ ¬ a ∈ st → st -- a = st;