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Proof step 3 from proof less-03 (previous step)

m < (m - n1) + n0, n0 ≤ n1, n0 ≤ m, n1 ≤ m ⊦ m + n0 = n1

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, n1, n] flags: <rotated> <all paths>

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);% ⊦ m = m + n ↔ n = 0; (s)% ⊦ m < n → n ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s) Used terms: m, n1, n