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Proof step 5 from proof blastN-len (previous step)

# x = n0 + n, ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n → # butlastn(n, x0) = n0 ⊦ # butlastn(n, x) = n0

User applied rule cut formula

And obtained the following new goals

Proof step 6:
# x = n0 + n, ∀ x0, n0, n. # x0 < # x → # x0 = n0 + n → # butlastn(n, x0) = n0 ⊦ x = [], # butlastn(n, x) = n0

Rule argument

fma: x = []

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x;%size-01: ⊦ # [] = 0; from specification list-data%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: x = []