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Proof step 34 from proof rmpos-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
m < # x0, a ∈ x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -1l n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ⊦ a = x0[m], x0 = [], a ∈ x0 -1l m, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: delpos-rec: ⊦ (a ' + x) -1l (n + 1) = a + (x -1l n);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2