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Proof step 13 from proof subset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# st0 < # st1, a ∈ st1, a ∈ st0, st1 -- a ⊆ st0 -- a, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, st0[a]] = ∅, st1[a] = st0[a], st1 = ∅, st1 ⊆ st0[a, st1[a]]

Rule argument

Additional information

Used simplifier rules: size-in: ⊦ a ∈ st → # st[a, d] = # st;