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Proof step 21 from proof keyset-subarray-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # subarray(kar, m, n) = n; (s)