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Proof step 1 from proof union-size-ls-in (conclusion)

⊦ a ∈ s0 ∧ a ∈ s1 → #(s0 ∪ s1) < # s0 + # s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s0 + # s1 ≤ #(s0 ∪ s1), a ∈ s1, a ∈ s0 ⊦ s1 = ∅, s0 = ∅, # s1 = 0, # s0 = 0

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat