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Proof step 18 from proof prefixn-prefixn (previous step)

# (m ≤ # x ⊃ firstn(m, x);x) < n ⊦ (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 19:
# (m ≤ # x ⊃ firstn(m, x);x) < n ⊦ x = ((n < m ⊃ n;m) ≤ # x ⊃ firstn((n < m ⊃ n;m), x);x), m ≤ # x

Rule argument

pos: right 1

Additional information

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