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Proof step 5 from proof subset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# st0 < # st1, a ∈ st1, st1[a, d] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, d] = ∅, st1 = ∅

Rule argument

Additional information

Used simplifier rules: size-in: ⊦ a ∈ st → # st[a, d] = # st;%empty: ⊦ # st = 0 ↔ st = ∅;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic