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Proof step 10 from proof append-eq-right (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
# ar0 = # ar1, n < # ar1, n < # ar1 + # ar → (ar0 + ar)[n] = (ar1 + ar)[n] ⊦ ar0[n] = ar1[n]

Rule argument

pos: left 3 terms: n abort: F discard: T computed: F

Additional information

Used terms: n