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Proof step 21 from proof append-assoc (previous step)

n < # x + # x0 ⊦ (x + x0)[n] = (n < # x ⊃ x[n];x0[n - # x])

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ n < # x → (x + y)[n] = x[n]; (s)