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Proof step 2 from proof extension (previous step)

# x = # y, ∀ n. n < # y → x[n] = y[n] ⊦ x = y

User applied rule structural induction

And obtained the following new goals

Proof step 3:
# x = #(a + y0), ∀ n. n < #(a + y0) → x[n] = (a + y0)[n], ∀ x. # x = # y0 ∧ (∀ n. n < # y0 → x[n] = y0[n]) → x = y0 ⊦ x = a + y0

Rule argument

indvar: y

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)