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Proof step 10 from proof in-prefixn-in-shorter-prefix (previous step)

a ∈ (n0 ≤ # x ⊃ firstn(n0, x);x), n0 ≤ n1 ⊦ a ∈ x, n1 ≤ # x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# x < n1, a ∈ (n0 ≤ # x ⊃ firstn(n0, x);x), n0 ≤ n1 ⊦ a ∈ x

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)