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Proof step 9 from proof blastN-len (previous step)

# x = n0 + n, x ≠ [] → (∃ y, a. x = y + a '), ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n → # butlastn(n, x0) = n0 ⊦ # butlastn(n, x) = n0, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# y + 1 = n0 + n, ∀ x0, n0, n. # x0 < #(y + a ') → # x0 = n0 + n → # butlastn(n, x0) = n0 ⊦ # butlastn(n, y + a ') = n0

Rule argument

Additional information

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