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Proof step 3 from proof pls (previous step)

m < m0 ∧ n + 1 < n0 → (¬ m + n + 1 < m0 + n0 ↔ false), m < m0, n < n0 ⊦ m + n + 1 < m0 + n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < n0, m < m0 ⊦ n0 = 0, m0 = 0, n + 1 < n0, m + n + 1 < m0 + n0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%s: ⊦ m < n + 1 ↔ ¬ n < m;%ref-1: ⊦ ¬ n < n; from specification nat-basic1%lf-06: ⊦ ¬ m < n0 ∧ n < n0 → n < m; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1