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Proof step 5 from proof subarray-append-consec (previous step)

m0 + n0 + n1 ≤ # ar ⊦ #(subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)) = # subarray(ar, m0, n0 + n1) ∧ (∀ n. n < #(subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)) → (subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1))[n] = subarray(ar, m0, n0 + n1)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 7:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 8:
n < n0 + n1, m0 + n0 + n1 ≤ # ar ⊦ (subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1))[n] = ar[m0 + n]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (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(ar, m, n) = n; (s)