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Proof step 22 from proof prefixn-eq-prefixn-app (previous step)

# y = n1, x0 + x1 = y + z, # x0 < n0, n1 ≤ # x0 + # x1, n1 ≤ #(x0 + x1), # x0 ≤ n1 ∧ x1 = [] ∨ n1 = # x0 ∧ x1 ≠ [] ⊦ x0 = y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
x0 + x1 = y + z, # x0 < n0, # y ≤ # x0 + # x1, # x0 ≤ # y ∧ x1 = [] ∨ # y = # x0 ∧ x1 ≠ [] ⊦ x0 = y

Rule argument

Additional information

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