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Proof step 22 from proof rmpos-in (previous step)

n = m + 1, x0[m] ∈ x0, x0[m] ∈ (a0 ' + x0) -p n, n ≤ # x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -p n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ⊦ n = 0, x0 = [], dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
m < # x0, x0[m] ∈ x0 -p m, x0[m] ∈ x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -p n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ⊦ x0 = [], dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: delpos-rec: ⊦ (a ' + x) -p (n + 1) = a + (x -p n);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)% ⊦ 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)