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Proof step 20 from proof prefixn-prefixn (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ ¬ n < n; (s)