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Proof step 19 from proof append-assoc (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 20:
n < # x + # x0 ⊦ (x + x0)[n] = (n < # x ⊃ x[n];(n < # x0 + # x ⊃ x0[n - # x];nf(n - # x - # x0)))
Proof step 22:
¬ n < # x + # x0 ⊦ nf(n - # x + # x0) = (n < # x ⊃ x[n];(n < # x0 + # x ⊃ x0[n - # x];nf(n - # x - # x0)))

Rule argument

pos: right 1

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