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Proof step 1 from proof union-size-not-less-02 (conclusion)

⊦ # s0 ≤ m ∧ # s1 ≤ n → ¬ m + n + 1 < #(s0 ∪ s1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s1 ≤ n, # s0 ≤ m, m + n + 1 < #(s0 ∪ s1) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1