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Proof step 3 from proof restn-last (previous step)

n < #(a + x0), ∀ n. n < # x0 → restn(n, x0).last = x0.last ⊦ restn(n, a + x0).last = (a + x0).last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).last = x0.last ⊦ restn(n, a ' + x0).last = (a ' + x0).last

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: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)