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Proof step 6 from proof union-size-not-less-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
m + n < # s0 + # s1, m + n < #(s0 ∪ s1), # s1 ≤ n, # s0 ≤ m, #(s0 ∪ s1) ≤ # s0 + # s1 ⊦ m + n < m + n

Rule argument

fma: m + n < m + n

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: m + n < m + n