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Proof step 11 from proof min-le-diff-third (previous step)

⊦ n2 ≤ n2 - (n0 < (n1 < n2 ⊃ n1;n2) ⊃ n0;(n1 < n2 ⊃ n1;n2)) ↔ n0 = 0, n1 = 0, n2 = 0

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 12:
n2 ≤ n2 - (n0 < (n1 < n2 ⊃ n1;n2) ⊃ n0;(n1 < n2 ⊃ n1;n2)) ⊦ n2 ≤ n2 - (n0 < (n1 < n2 ⊃ n1;n2) ⊃ n0;(n1 < n2 ⊃ n1;n2)) ↔ n0 = 0, n1 = 0, n2 = 0

Rule argument

fma: n2 ≤ n2 - (n0 < (n1 < n2 ⊃ n1;n2) ⊃ n0;(n1 < n2 ⊃ n1;n2))

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m - 0 = m; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)% ⊦ 0 ≤ m; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: n2 ≤ n2 - (n0 < (n1 < n2 ⊃ n1;n2) ⊃ n0;(n1 < n2 ⊃ n1;n2))