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Proof step 4 from proof pls (previous step)

n < n0, m < m0 ⊦ n0 = 0, m0 = 0, n + 1 < n0, m + n + 1 < m0 + n0

User applied rule structural induction

And closed the goal!

Rule argument

indvar: n0

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%oz: ⊦ m + 1 ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%s: ⊦ m < n + 1 ↔ ¬ n < m;%l: ⊦ m + n < m + n0 ↔ n < n0;%nat freely generated by 0, +1;;%ls-03: ⊦ ¬ m < n → (¬ n < m ↔ n = m); from specification nat-basic1%lf-05: ⊦ ¬ n < m ∧ m = n0 → ¬ n < n0; (forward) from specification nat-basic1%lf-04: ⊦ ¬ n < m ∧ n = n0 → ¬ n0 < m; (forward) from specification nat-basic1