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Proof step 7 from proof lastN-blastN (previous step)

n ≤ # x, x ≠ [] → x = y + a ', ∀ x0, n. # x0 < # x → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ x = [], butlastn(n, x) + lastn(n, x) = x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ≤ # y + 1, ∀ x0, n. # x0 < #(y + a ') → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ butlastn(n, y + a ') + lastn(n, y + a ') = y + a '

Rule argument

Additional information

Used simplifier rules: len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%e-02: ⊦ x + a ' ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list