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Proof step 18 from proof append-assoc (previous step)

⊦ ⌜λ n. (n < #(x + x0) ⊃ (x + x0)[n];nf(n - #(x + x0)))⌝ = ⌜λ n. (n < # x ⊃ x[n];(n < # x0 + # x ⊃ x0[n - # x];nf(n - # x - # x0)))⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ (n < # x + # x0 ⊃ (x + x0)[n];nf(n - # x + # x0)) = (n < # x ⊃ x[n];(n < # x0 + # x ⊃ x0[n - # x];nf(n - # x - # x0)))

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)