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Proof step 7 from proof in-prefixn-pos-smaller (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ firstn(n1, x0), n1 ≤ pos(a, x0), ∀ a1, n0. ¬ a1 ∈ firstn(n0, x0) ∨ ¬ n0 ≤ pos(a1, x0) ⊦ a = a0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1; (s)% ⊦ ¬ m < 0; (s)% ⊦ pos(a, a ' + x) = 0; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)% ⊦ firstn(n + 1, a ' + x) = a + firstn(n, x); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)