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Proof step 1 from proof sub-ls-rew (conclusion)

m0 < m, n0 < n ⊦ m - m0 < n - n0 ↔ m + n0 < n + m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m0 < m, n0 < n ⊦ m < (n - n0) + m0 ↔ m + n0 < n + m0, m = 0, n = 0

Rule argument

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n;%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1