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Proof step 8 from proof append-inj (previous step)

# ar0 = # ar1, # ar = # ar2, ∀ n. n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n] ⊦ ar0 = ar1 ∧ ar = ar2

User applied rule case distinction

And obtained the following new goals

Proof step 9:
# ar = # ar2, # ar0 = # ar1, ∀ n. n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n] ⊦ ar0 = ar1
Proof step 18:
# ar = # ar2, # ar0 = # ar1, ∀ n. n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n] ⊦ ar = ar2

Rule argument

pos: right 1

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