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

# ar = n0 + n, n0 = 0, m ≤ n0 ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: append-zero-left: ⊦ mkarray(0) + ar = ar;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%subarray-elim-right: n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m);% ⊦ # ar = 0 → (mkarray(0) = ar ↔ true); (ss)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m - 0 = m; (s)% ⊦ # mkarray(n) = n; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ subarray(ar, n, 0) = mkarray(0); (s)% ⊦ m ≤ 0 ↔ m = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)