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Proof step 1 from proof elim-lastn-butlastn (conclusion)

⊦ n ≤ # x → (y = butlastn(n, x) ∧ z = lastn(n, x) ↔ x = y + z ∧ # z = n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ y = butlastn(# z, y + z) ∧ z = lastn(# z, y + z)

Rule argument

Additional information

Used simplifier rules: lastN-blastN: ⊦ n ≤ # x → butlastn(n, x) + lastn(n, x) = x;%lastn-length-01: ⊦ n ≤ # x → # lastn(n, x) = n;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list