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Proof step 4 from proof isprefix-app-prefixn-isprefix (previous step)

x0 ⊑ x1 + (n ≤ # x2 ⊃ firstn(n, x2);x2) ⊦ x0 ⊑ x1 + x2

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
n ≤ # x2, x0 ⊑ x1 + firstn(n, x2) ⊦ x0 ⊑ x1 + x2

Rule argument

pos: left 1

Additional information