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Proof step 3 from proof prefixn-size-prefixn (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
# (n ≤ # x ⊃ firstn(n, x);x) ≤ # x ⊦ firstn(# (n ≤ # x ⊃ firstn(n, x);x), x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 6:
⊦ x = (n ≤ # x ⊃ firstn(n, x);x), # (n ≤ # x ⊃ firstn(n, x);x) ≤ # x

Rule argument

pos: right 1

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