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Proof step 4 from proof rmpos-in (previous step)

n < #(a0 + x0), ∀ n, a. n < # x0 → (a ∈ x0 -p n ↔ a ≠ x0[n] ∧ a ∈ x0) ∨ x0 = [] ∨ dups(x0) ⊦ a0 + x0 = [], a ∈ (a0 + x0) -p n ↔ a ≠ (a0 + x0)[n] ∧ a ∈ a0 + x0, dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ # x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -p n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ∨ x0 = [] ⊦ a ∈ (a0 ' + x0) -p n ↔ a ≠ (a0 ' + x0)[n] ∧ (a = a0 ∨ a ∈ x0), dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)