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Proof step 8 from proof in-prefixn-pos-smaller0 (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n1, pos(a, x0), a

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) Used terms: n1, pos(a, x0), a