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Proof step 6 from proof pos-get (previous step)

¬ n < # a ', n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ pos(x0[n - # a '], a ' + x0) = n, a ∈ x0, dups(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ pos(x0[n - 1], a ' + x0) = n, n = 0, a ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%len-02: ⊦ # a ' = 0 +1; from specification list