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Proof step 15 from proof isprefix-add-prefixn-size (previous step)

#(x0 + z) = #(x1 + x2), # x1 + n < # x0, # x2 < n, ∀ n. n < #(x0 + z) → (x0 + z)[n] = (x1 + x2)[n] ⊦ x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# x0 + # z = # x1 + # x2, # x2 < n, # x1 + n < # x0, ∀ n0. n0 < # x1 + # x2 → (x0 + z)[n0] = (x1 + x2)[n0] ⊦ x0 = []

Rule argument

Additional information

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