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Proof step 3 from proof diffone-nodupsget (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: diff-firstofapp: ⊦ (x + y) \ x = y \ x;%diff-onenonmember: ⊦ ¬ a ∈ x → x \ a ' = x;%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1; from specification list-del%get-nodups: ⊦ m < # x ∧ n < # x ∧ ¬ dups(x) → (x[m] = x[n] ↔ m = n); from specification list-del%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%pos-y: ⊦ pos(a, a ' + x) = 0; from specification list-del%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%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%get-succ: ⊦ (a ' + x)[n + 1] = x[n]; from specification list-del%c: ⊦ a + x = a ' + x; from specification list