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Proof step 4 from proof append-eq-right (previous step)

#(ar0 + ar) = #(ar1 + ar), ∀ n. n < #(ar0 + ar) → (ar0 + ar)[n] = (ar1 + ar)[n] ⊦ ar0 = ar1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 6:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 7:
# ar0 = # ar1, ∀ n. n < # ar1 + # ar → (ar0 + ar)[n] = (ar1 + ar)[n] ⊦ ar0 = ar1

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 = m + n0 ↔ n = n0; (s)