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Proof step 5 from proof keyset-subarray-eq (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 6:
subarray(kar0, 0, m0 + 1) = subarray(kar1, 0, m0 + 1), m0 < # kar0, m0 < # kar1, subarray(kar0, 0, m0) = subarray(kar1, 0, m0) ∧ m0 ≤ # kar0 ∧ m0 ≤ # kar1 → keyset(kar0, m0) = keyset(kar1, m0) ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0]

Rule argument

pos: left 4 terms: kar0, kar1 abort: F discard: T computed: F

Additional information

Used terms: kar0, kar1