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Proof step 2 from proof subarray-eq-elim (previous step)

n0 = # ar, n + m ≤ # ar ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = n0 - n + m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 4:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 5:
n + m ≤ # ar ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n - n + n1 = m - n1; (s)