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Proof step 2 from proof prefixn-size-eq (previous step)

# x0 = # x1 ⊦ # prefixn(n, x0) = # prefixn(n, x1) ↔ true

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

Rule argument

Additional information

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