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Proof step 3 from proof rmpos-in (previous step)

∀ n, a. n < # x0 → (a ∈ x0 -1l n ↔ a ≠ x0[n] ∧ a ∈ x0) ∨ x0 = [] ∨ dups(x0), n < #(a0 + x0) ⊦ a ∈ (a0 + x0) -1l n ↔ a ≠ (a0 + x0)[n] ∧ a ∈ a0 + x0, a0 + x0 = [], dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -1l n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ∨ x0 = [] ⊦ a ∈ (a0 ' + x0) -1l n ↔ a ≠ (a0 ' + x0)[n] ∧ (a = a0 ∨ a ∈ x0), a0 ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list