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Proof step 2 from proof subset-size (previous step)

∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ # st0 < # st1[a, d] → ¬ st1[a, d] ⊆ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# st0 < # st1[a, d], st1[a, d] ⊆ st0, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a, d] = ∅

Rule argument

Additional information

Used simplifier rules: empty: ⊦ # st = 0 ↔ st = ∅;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1