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Proof step 5 from proof union-size-not-less-01 (previous step)

m + n < #(s0 ∪ s1), # s1 ≤ n, # s0 ≤ m, #(s0 ∪ s1) ≤ # s0 + # s1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m + n < #(s0 ∪ s1), m + n < # s0 + # s1, #(s0 ∪ s1) ≤ # s0 + # s1, # s0 ≤ m, # s1 ≤ n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat