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Proof step 4 from proof min-notls2 (previous step)

m < (n < m ⊃ n;m) ⊦ n = 0

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → ¬ n < m; (s)