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Proof step 3 from proof pos-get (previous step)

∀ n. n < # x0 → pos(x0[n], x0) = n ∨ x0 = [] ∨ dups(x0), n < #(a + x0) ⊦ pos((a + x0)[n], a + x0) = n, a + x0 = [], dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ pos((a ' + x0)[n], a ' + x0) = n, a ∈ 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%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list