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Proof step 4 from proof diffone-nodupsget (previous step)

n + 1 < #(a0 + x0), a ∈ a0 + x0, ∀ n, a. n + 1 < # x0 ∧ a ∈ x0 → (x0 \ a ')[n] = (n < pos(a, x0) ⊃ x0[n];x0[n + 1]) ∨ x0 = [] ∨ dups(x0) ⊦ a0 + x0 = [], ((a0 + x0) \ a ')[n] = (n < pos(a, a0 + x0) ⊃ (a0 + x0)[n];(a0 + x0)[n + 1]), dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < # x0, a ∈ x0, ∀ n0, a1. n0 + 1 < # x0 ∧ a1 ∈ x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ ((a0 ' + x0) \ a ')[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), x0 = [], dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: diff-firstofapp: ⊦ (x + y) \ x = y \ x;%diff-onenonmember: ⊦ ¬ a ∈ x → x \ a ' = x;% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1; (s)% ⊦ m < # x ∧ n < # x ∧ ¬ dups(x) → (x[m] = x[n] ↔ m = n); (s)% ⊦ ¬ m < 0; (s)% ⊦ pos(a, a ' + x) = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ (a ' + x)[n + 1] = x[n]; (s)% ⊦ a + x = a ' + x; (s)