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Proof step 6 from proof append-mod (previous step)

n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → x0 = [] ∨ ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ # x2 + # x0 = 0 ⊦ x0 = [], (x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 7:
x0[n0, a] = x1, n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → x0 = [] ∨ ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ # x2 + # x0 = 0 ⊦ x0 = [], (x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1
Proof step 9:
n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → x0 = [] ∨ ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ # x2 + # x0 = 0 ⊦ x0[n0, a] = x1, x0 = [], (x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1

Rule argument

fma: x0[n0, a] = x1

Additional information

Used formulas: x0[n0, a] = x1