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Proof step 8 from proof sublist-lenrec-01 (previous step)

y = a0 + y0, n0 < # y, ∀ y, a. n0 ≤ # y → a ' + firstn(n0, y) = firstn(n0, a ' + y) + (a ' + y)[n0] ' ⊦ a0 ' + firstn(n0, y0) = firstn(n0, y) + y[n0] ', y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n0 ≤ # y0, ∀ y, a. n0 ≤ # y → a ' + firstn(n0, y) = firstn(n0, a ' + y) + (a ' + y)[n0] ' ⊦ a0 ' + firstn(n0, y0) = firstn(n0, a0 ' + y0) + (a0 ' + y0)[n0] '

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)