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Proof step 4 from proof min-le-left (previous step)

n0 ≤ n1 ⊦ (n0 < n1 ⊃ n0;n1) = n0

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)