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Proof step 8 from proof firstn-last-rec (previous step)

y = a + y0, m < # y, ∀ x0. m < # x0 → x0.first ' + firstn(m, x0.rest) = firstn(m, x0) + x0[m] ' ⊦ a ' + firstn(m, y0) = firstn(m, y) + y[m] ', y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m ≤ # y0, ∀ x0. m < # x0 → x0.first ' + firstn(m, x0.rest) = firstn(m, x0) + x0[m] ' ⊦ a ' + firstn(m, y0) = firstn(m, a ' + y0) + (a ' + y0)[m] '

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list