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Proof step 4 from proof prefixn-shift-right (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
# x0 ≤ n, n ≤ # x0 + # x1 ⊦ firstn(n, x0 + x1) = x0 + (n ≤ # x1 + # x0 ⊃ firstn(n - # x0, x1);x1)

Rule argument

pos: right 1

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list