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Proof step 6 from proof blastN-rec-01 (previous step)

n ≤ # x, x ≠ [] → x = y + a0 ', ∀ x0, n, a. # x0 < # x → n ≤ # x0 → butlastn(n, a ' + x0) = a ' + butlastn(n, x0) ⊦ butlastn(n, a ' + x) = a ' + butlastn(n, x)

User applied rule cut formula

And obtained the following new goals

Proof step 7:
n ≤ # x, x ≠ [] → x = y + a0 ', ∀ x0, n, a. # x0 < # x → n ≤ # x0 → butlastn(n, a ' + x0) = a ' + butlastn(n, x0) ⊦ x = [], butlastn(n, a ' + x) = a ' + butlastn(n, x)

Rule argument

fma: x = []

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x;%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%size-01: ⊦ # [] = 0; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%c-04: ⊦ a ' = b ' ↔ a = b; from specification list Used formulas: x = []