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Proof step 2 from proof union-size-not-less-01 (previous step)

# s1 ≤ n, # s0 ≤ m, m + n < #(s0 ∪ s1) ⊦ #(s0 ∪ s1) = 0, s0 ∪ s1 = ∅

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: set-union inst: name: union-size-le seq: ⊦ #(s0 ∪ s1) ≤ # s0 + # s1 subst: [s0, s1] → [s0, s1] flags: current, []

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%ls-add-not-01: ⊦ m ≤ m0 ∧ n ≤ n0 → ¬ m0 + n0 < m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: s0, s1