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Proof step 7 from proof firstn-firstn (previous step)

firstn(n, a ' + y) = a0 + y0 ∧ firstn(n, a ' + y) ≠ [], m0 ≤ # y, n ≤ # y + 1, m0 < n, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(m0, y0) = firstn(m0, y), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
firstn(n, a ' + y) = a0 ' + y0, m0 ≤ # y, n ≤ # y + 1, m0 < n, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(m0, y0) = firstn(m0, y), n = 0

Rule argument

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%c: ⊦ a + x = a ' + x; from specification list