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Proof step 3 from proof extension (previous step)

# 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x = 1 + # y0, ∀ x. # x = # y0 ∧ (∀ n. n < # y0 → x[n] = y0[n]) → x = y0, ∀ n. ¬ # y0 < n → x[n] = (a ' + y0)[n] ⊦ x = a ' + y0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)