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Proof step 14 from proof keyset-subarray-eq (previous step)

n < # kar1, n < # kar0, n < m0, m0 < # kar1, m0 < # kar0, n + 1 < # kar1, n + 1 < # kar0, ∀ n0. ¬ m0 < n0 → subarray(kar0, 0, m0 + 1)[n0] = subarray(kar1, 0, m0 + 1)[n0] ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0], kar0[n] = kar1[n]

User applied rule all left

And obtained the following new goals

Proof step 15:
n + 1 < # kar0, n + 1 < # kar1, m0 < # kar0, m0 < # kar1, n < m0, n < # kar0, n < # kar1, ¬ m0 < n → subarray(kar0, 0, m0 + 1)[n] = subarray(kar1, 0, m0 + 1)[n] ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0], kar0[n] = kar1[n]

Rule argument

pos: left 8 terms: n abort: F discard: T computed: F

Additional information

Used terms: n