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Proof step 12 from proof lastN-blastN (previous step)

n = m + 1, n ≤ # y + 1, ∀ x0, n. # x0 < #(y + a ') → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ butlastn(n, y + a ') + lastn(m, y) = y, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m ≤ # y, ∀ x0, n. # x0 < #(y + a ') → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ butlastn(m, y) + lastn(m, y) = y

Rule argument

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%butlast-def: ⊦ (x + a ').butlast = x; from specification list-last%butlastN-rec: ⊦ butlastn(n + 1, x) = butlastn(n, x.butlast); from specification list-last