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Proof step 2 from proof union-size-not-less-02 (previous step)

# s1 ≤ n, # s0 ≤ m, m + n + 1 < #(s0 ∪ s1) ⊦

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: set-union inst: name: union-size-not-less-01 seq: ⊦ # s0 ≤ m ∧ # s1 ≤ n → ¬ m + n < #(s0 ∪ s1) subst: [s0, m, s1, n] → [s0, m, s1, n + 1] flags: current, []

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%l-02: ⊦ ¬ m + n < m; from specification nat-basic2 Used terms: s0, m, s1, n + 1