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Proof step 19 from proof get-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
x0[m1] = x0[m0], m1 < # x0, m0 < # x0, ∀ m, n0. x0[m] = x0[n0] ∧ m < # x0 ∧ n0 < # x0 → m = n0 ⊦ m1 = m0, x0 = [], dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)