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Proof step 1 from proof le-num-02 (conclusion)
n0 < n ⊦ m + (* n) ≤ m0 + (* n0) ↔ m + (n - n0) ≤ m0
Special step normalize
And obtained the following new goals
Proof step 2
:
n0 < n ⊦ m + n ≤ m0 + n0 ↔ m + (n - n0) ≤ m0