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Proof step 18 from proof min-le-diff-third (previous step)

n2 = n3 + (n1 < n2 ⊃ n1;n2), (n1 < n2 ⊃ n1;n2) ≤ n2, n2 ≤ n3, n1 ≤ n0, (n1 < n2 ⊃ n1;n2) ≤ n3 ⊦ 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 = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → m ≤ n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s) Used formulas: n1 < n2