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Proof step 2 from proof twoeqf (previous step)

m + m0 = 2 * n, n < m, n ≤ m0 ⊦ m = 0

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
m ≤ m0, n < n0 ⊦ m + n < m0 + n0

Rule argument

spec: nat inst: name: pls seq: m ≤ m0, n < n0 ⊦ m + n < m0 + n0 subst: [m, m0, n, n0] → [n, m0, n, m] flags:

Additional information

Used simplifier rules: two-02: ⊦ m + m = 2 * m;%two-zero-01: ⊦ 2 * m = 0 ↔ m = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: n, m0, n, m