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Proof step 6 from proof in-prefixn-pos-smaller0 (previous step)

n < pos(a, a0 ' + x0), n < m, a ∈ firstn(n, a0 ' + x0), m ≤ pos(a, a0 ' + x0), ∀ n0, m0, a1. ¬ n0 < m0 ∨ ¬ n0 < pos(a1, x0) ∨ ¬ a1 ∈ firstn(n0, x0) ∨ ¬ m0 ≤ pos(a1, x0) ⊦

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
n = n1 +1, n < m, n < pos(a, a0 ' + x0), a ∈ firstn(n, a0 ' + x0), m ≤ pos(a, a0 ' + x0), ∀ n0, m0, a1. ¬ n0 < m0 ∨ ¬ n0 < pos(a1, x0) ∨ ¬ a1 ∈ firstn(n0, x0) ∨ ¬ m0 ≤ pos(a1, x0) ⊦

Rule argument

var: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ firstn(0, x) = []; (s)% ⊦ ¬ a ∈ []; (s)