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Proof step 6 from proof ls-add-not (previous step)

m + n ≤ m0 + n0, m0 + n ≤ m0 + n0, m + n ≤ m0 + n, m0 + n0 + n1 < m + n, m ≤ m0, n ≤ n0 ⊦ m0 + n0 + n1 < m0 + n0, m + n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: trans-ls-le seq: m < n, n ≤ n0 ⊦ m < n0 subst: [m, n, n0] → [m0 + n0 + n1, m + n, m0 + n0] flags: rewrite, current

Additional information

Used simplifier rules: le-sum: ⊦ n ≤ n0 ∧ n1 ≤ m → n + n1 ≤ n0 + m;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%l-04: ⊦ m + m0 < n + n0 → (¬ m + m0 < n + n0 ↔ false); from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m0 + n0 + n1, m + n, m0 + n0