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Proof step 24 from proof keyset-subarray-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: array-copy, inst: key-index-array name: subarray-at seq: ⊦ n0 < n → subarray(kar, m, n)[n0] = kar[m + n0] subst: [n0, n, m, kar] → [m0, m0 + 1, 0, kar0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n0 < n → subarray(kar, m, n)[n0] = kar[m + n0]; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%# kar = n ⊦ subarray(kar, 0, n) = kar; (s) Used terms: m0, m0 + 1, 0, kar0