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Proof step 10 from proof keyset-subarray-eq (previous step)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
subarray(kar0, 0, m0 + 1) = subarray(kar1, 0, m0 + 1), n < # kar1, n < # kar0, n < m0, m0 < # kar1, m0 < # kar0, n + 1 < # kar1, n + 1 < # kar0 ⊦ keyset(kar0, m0) ++ kar0[m0] = keyset(kar1, m0) ++ kar1[m0], kar0[n] = kar1[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(kar, m, n)[n0] = kar[m + n0]; (s)% ⊦ # subarray(kar, m, n) = n; (s)