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Proof step 2 from proof isprefix-app-prefixn-isprefix (previous step)

x0 ⊑ x1 + prefixn(n, x2) ⊦ x0 ⊑ x1 + x2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 4:
x0 ⊑ x1 + (n ≤ # x2 ⊃ firstn(n, x2);x2) ⊦ x0 ⊑ x1 + x2

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);