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Proof step 14 from proof get-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
x0[m1] = (a ' + x0)[n], m1 < # x0, n ≤ # x0, ∀ m, n0. x0[m] = x0[n0] ∧ m < # x0 ∧ n0 < # x0 → m = n0 ⊦ m1 + 1 = n, x0 = [], dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (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)