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Proof step 9 from proof firstn-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m ≤ # y0, ∀ y, a. m ≤ # y → (a ' + firstn(m, y)).last = (a ' + y)[m] ⊦ (a0 ' + firstn(m, y0)).last = (a0 ' + 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