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Proof step 5 from proof mkemparray-append (previous step)

⊦ #(fill(mkarray(n0), empty) + fill(mkarray(n1), empty)) = # fill(mkarray(n0 + n1), empty) ∧ (∀ n. n < #(fill(mkarray(n0), empty) + fill(mkarray(n1), empty)) → (fill(mkarray(n0), empty) + fill(mkarray(n1), empty))[n] = fill(mkarray(n0 + n1), empty)[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 ⊦ (fill(mkarray(n0), empty) + fill(mkarray(n1), empty))[n] = empty

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% ⊦ m < n → fill(mkarray(n), a)[m] = a; (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)% ⊦ # mkarray(n) = n; (s)% ⊦ # fill(ar, a) = # ar; (s)