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Proof step 7 from proof get-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a = (a ' + x0)[n], n ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ m0 < # x0 ∧ n0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ 0 = n, dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;% ⊦ 0 ≤ m; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list