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Proof step 2 from proof violations-swap (previous step)

x[n + 1] < x[n], n + 1 < # x ⊦ violations(x[n, x[n + 1]][n + 1, x[n]]) = violations(x) - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x[n + 1] < x[n], n + 1 < # x ⊦ violations(x[n, x[n + 1]][n + 1, x[n]]) = violations(x) - 1, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic