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Proof step 7 from proof extension (previous step)

# x = 1 + # y0, x = a0 + y, # y = # y0 ∧ (n < # y0 → y[n] = y0[n]) → y = y0, ∀ n. ¬ # y0 < n → x[n] = (a ' + y0)[n] ⊦ x = a ' + y0, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# y = # y0, (n < # y0 → y[n] = y0[n]) → y = y0, ∀ n0. ¬ # y0 < n0 → (a0 ' + y)[n0] = (a ' + y0)[n0] ⊦ a0 ' + y = a ' + y0

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%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)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)