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Proof step 1 from proof max-eq-0 (conclusion)

⊦ max(n, m) = 0 ↔ n = 0 ∧ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ max(n, m) = (n < m ⊃ m;n)
Proof step 3:
⊦ m ≤ n → max(n, m) = n
Proof step 4:
(n < m ⊃ m;n) = 0 ⊦ n = 0 ∧ m = 0

Rule argument

Additional information

Used simplifier rules: max: ⊦ max(n, m) = (n < m ⊃ m;n);%max-second-le: ⊦ m ≤ n → max(n, m) = n;