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Proof step 1 from proof sizesubl (conclusion)

⊦ s ⊆ s0 ∧ a ∈ s0 ∧ ¬ a ∈ s → # s < # s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s0 ≤ # s, a ∈ s0, s ⊆ s0 ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat