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Proof step 1 from proof union-size-01 (conclusion)

⊦ ¬ a ∈ s → ¬ #((s ∪ s0) ++ a) ≤ # s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: union-sizele-01: ⊦ #((s ∪ s0) ++ a) ≤ # s ↔ s0 ⊆ s ∧ a ∈ s;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;