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Proof step 3 from proof neq-01 (previous step)

n0 + 1 = n + m0, n + m ≤ n0, m0 ≤ m ⊦

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;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ n + 0 = n; (s)