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Proof step 9 from proof isprefix-add-prefixn-size (previous step)

# y = n, x0 + z = x1 + y, x2 = y + z0, # x1 + n < # x0, n ≤ # x2 ⊦ x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
x0 + z = x1 + y, # x1 + # y < # x0 ⊦ x0 = []

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 ≤ m + n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)