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Proof step 7 from proof extension (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list