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Proof step 4 from proof prefixn-size-prefixn (previous step)

# (n ≤ # x ⊃ firstn(n, x);x) ≤ # x ⊦ firstn(# (n ≤ # x ⊃ firstn(n, x);x), x) = (n ≤ # x ⊃ firstn(n, x);x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ firstn(# (n ≤ # x ⊃ firstn(n, x);x), x) = (n ≤ # x ⊃ firstn(n, x);x)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ n ≤ # x → # firstn(n, x) = n; (s)