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Proof step 18 from proof get-nodups (previous step)

n = m0 + 1, x0[m1] = x0[m0], m1 < # x0, n ≤ # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ m1 + 1 = n, x0 = [], n = 0, dups(x0), a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
x0[m1] = x0[m0], m1 < # x0, m0 < # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ x0 = [], m1 = m0, dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2