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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
n ≤ # (m ≤ # x ⊃ firstn(m, x);x) ⊦ firstn(n, (m ≤ # x ⊃ firstn(m, x);x)) = ((n < m ⊃ n;m) ≤ # x ⊃ firstn((n < m ⊃ n;m), x);x)
Proof step 17:
⊦ (m ≤ # x ⊃ firstn(m, x);x) = ((n < m ⊃ n;m) ≤ # x ⊃ firstn((n < m ⊃ n;m), x);x), n ≤ # (m ≤ # x ⊃ firstn(m, x);x)

Rule argument

pos: right 1

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information