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Proof step 18 from proof dups-def (previous step)

m = m0 + 1 ∧ m ≠ 0, (a ' + x0)[m] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0 ⊦ m = n, a ∈ x0, dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], m0 < # x0 ∧ n - 1 < # x0 ∧ m0 ≠ n - 1 ∧ x0[m0] = x0[n - 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
x0[m0] = (a ' + x0)[n], m0 < # x0, n ≤ # x0 ⊦ m0 + 1 = n, x0 = [], dups(x0), a ∈ x0, n - 1 < # x0 ∧ m0 ≠ n - 1 ∧ x0[m0] = x0[n - 1], ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2