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Proof step 2 from proof neq-01 (previous step)

n0 + 1 = n + m0, m0 ≤ m, n + m ≤ n0 ⊦

User applied rule structural induction

And closed the goal!

Rule argument

indvar: n

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%succ-01: ⊦ m + 1 ≤ n ↔ m < n;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2