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Proof step 18 from proof subset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# st0 ≤ # st1, st1[a, d] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, d] = ∅, a ∈ st1

Rule argument

Additional information

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