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Proof step 2 from proof sub-ls-rew (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < n, m0 < m ⊦ m < (n - n0) + m0 ↔ m + n0 < n + m0

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ m ≤ n ↔ ¬ n < m;%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)