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Proof step 8 from proof keyset-subarray-eq (previous step)

subarray(kar0, 0, m0 + 1) = subarray(kar1, 0, m0 + 1), m0 < # kar0, m0 < # kar1 ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0], subarray(kar0, 0, m0) = subarray(kar1, 0, m0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ kar = kar0 ↔ # kar = # kar0 ∧ (∀ n. n < # kar → kar[n] = kar0[n])
Proof step 10:
subarray(kar0, 0, m0 + 1) = subarray(kar1, 0, m0 + 1), m0 < # kar1, m0 < # kar0 ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0], # subarray(kar0, 0, m0) = # subarray(kar1, 0, m0) ∧ (∀ n. n < # subarray(kar0, 0, m0) → subarray(kar0, 0, m0)[n] = subarray(kar1, 0, m0)[n])

Rule argument

spec: array, inst: key-index-array name: Extension seq: ⊦ kar = kar0 ↔ # kar = # kar0 ∧ (∀ n. n < # kar → kar[n] = kar0[n]) subst: [kar0, kar] → [subarray(kar1, 0, m0), subarray(kar0, 0, m0)] flags: <all paths>

Additional information

Used terms: subarray(kar1, 0, m0), subarray(kar0, 0, m0)