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Proof step 3 from proof get-butlast (previous step)

∀ n, a. n < # x0 → (x0 + a ')[n] = x0[n] ∨ x0 = [], n < #(a0 + x0) ⊦ ((a0 + x0) + a ')[n] = (a0 + x0)[n], a0 + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # x0, ∀ n, a. n < # x0 → (x0 + a ')[n] = x0[n] ∨ x0 = [] ⊦ ((a0 ' + x0) + a ')[n] = (a0 ' + x0)[n]

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%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list