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Proof step 5 from proof firstn-restn-eq (previous step)

x = a + y, n0 < # x, ∀ x. n0 ≤ # x → firstn(n0, x) + restn(n0, x) = x ⊦ (a ' + firstn(n0, y)) + restn(n0, y) = x, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c: ⊦ a + x = a ' + x; from specification list