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Proof step 14 from proof min-le-diff-third (previous step)

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

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: n1 < n2

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → ¬ n < 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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]% ⊦ m +1 = m + 1; (s) Used formulas: n1 < n2