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Proof step 16 from proof min-le-diff-third (previous step)

n2 ≤ n2 - (n1 < n2 ⊃ n1;n2), n1 ≤ n0 ⊦ n0 = 0, n1 = 0, n2 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
n2 = n3 + (n1 < n2 ⊃ n1;n2), n1 ≤ n0, n2 ≤ n3, (n1 < n2 ⊃ n1;n2) ≤ n2 ⊦ n0 = 0, n1 = 0, n2 = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [(n1 < n2 ⊃ n1;n2), n2, n]

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ n < n0 → (n0 < n ↔ false); (ss)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: (n1 < n2 ⊃ n1;n2), n2, n