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Proof step 4 from proof put-firstn-restn (previous step)

x = a0 + y, n < # x ⊦ x[n, a] = firstn(n, x) + a ' + restn(n, y), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ # y ⊦ (a0 ' + y)[n, a] = firstn(n, a0 ' + y) + a ' + restn(n, y)

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)