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Proof step 5 from proof prefixn-shift-right (previous step)

# x0 ≤ n, n ≤ # x0 + # x1 ⊦ firstn(n, x0 + x1) = x0 + (n ≤ # x1 + # x0 ⊃ firstn(n - # x0, x1);x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# x0 ≤ n, n ≤ # x0 + # x1 ⊦ firstn(n, x0 + x1) = x0 + firstn(n - # x0, x1)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic