Proof step 1 from proof max-a (conclusion) (back to tree)

⊦ max(m, max(n, n0)) = max(max(m, n), n0)

System applied rule normalize

And obtained the following new goals

Proof step 2:
max(m, max(n, n0)) ≠ max(max(m, n), n0) ⊦

Rule argument

Additional information

Proof step 2 from proof max-a (previous step) (back to tree)

max(m, max(n, n0)) ≠ max(max(m, n), n0) ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
max(m, max(n, n0)) ≠ max(max(m, n), n0), m < max(n, n0) ⊦
Proof step 9:
max(m, max(n, n0)) ≠ max(max(m, n), n0), ¬ m < max(n, n0) ⊦

Rule argument

fma: m < max(n, n0)

Additional information

Used formulas: m < max(n, n0)

Proof step 3 from proof max-a (previous step) (back to tree)

max(m, max(n, n0)) ≠ max(max(m, n), n0), m < max(n, n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
max(n, n0) ≠ max(max(m, n), n0), m < max(n, n0) ⊦

Rule argument

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 4 from proof max-a (previous step) (back to tree)

max(n, n0) ≠ max(max(m, n), n0), m < max(n, n0) ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
max(n, n0) ≠ max(max(m, n), n0), m < max(n, n0), n < n0 ⊦

Rule argument

fma: n < n0

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; Max-right: ⊦ ¬ m < n → max(m, n) = m; Used formulas: n < n0

Proof step 5 from proof max-a (previous step) (back to tree)

max(n, n0) ≠ max(max(m, n), n0), m < max(n, n0), n < n0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 ≠ max(max(m, n), n0), m < n0, n < n0 ⊦

Rule argument

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 6 from proof max-a (previous step) (back to tree)

n0 ≠ max(max(m, n), n0), m < n0, n < n0 ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
n0 ≠ max(max(m, n), n0), n < n0, m < n0, ¬ max(m, n) < n0 ⊦

Rule argument

fma: max(m, n) < n0

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; Used formulas: max(m, n) < n0

Proof step 7 from proof max-a (previous step) (back to tree)

n0 ≠ max(max(m, n), n0), n < n0, m < n0, ¬ max(m, n) < n0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 ≠ max(m, n), m < max(m, n), n0 < max(m, n), n < n0, m < n0, n + 1 < max(m, n), n < max(m, n), m + 1 < max(m, n) ⊦

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss)

Proof step 8 from proof max-a (previous step) (back to tree)

n0 ≠ max(m, n), m < max(m, n), n0 < max(m, n), n < n0, m < n0, n + 1 < max(m, n), n < max(m, n), m + 1 < max(m, n) ⊦

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ ¬ n < n; (s) ⊦ m < n → ¬ n < m; (s) Used formulas: m < n

Proof step 9 from proof max-a (previous step) (back to tree)

max(m, max(n, n0)) ≠ max(max(m, n), n0), ¬ m < max(n, n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m ≠ max(max(m, n), n0), max(n, n0) ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 10 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), max(n, n0) ≤ m ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 11:
m ≠ max(max(m, n), n0), n < n0, max(n, n0) ≤ m ⊦
Proof step 17:
m ≠ max(max(m, n), n0), max(n, n0) ≤ m, ¬ n < n0 ⊦

Rule argument

fma: n < n0

Additional information

Used formulas: n < n0

Proof step 11 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), n < n0, max(n, n0) ≤ m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m ≠ max(max(m, n), n0), n < n0, n < m, n0 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 12 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), n < n0, n < m, n0 ≤ m ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 13:
m ≠ max(max(m, n), n0), n < m, n < n0, max(m, n) < n0, n0 ≤ m ⊦
Proof step 15:
m ≠ max(max(m, n), n0), n < m, n < n0, n0 ≤ m, ¬ max(m, n) < n0 ⊦

Rule argument

fma: max(m, n) < n0

Additional information

Used formulas: max(m, n) < n0

Proof step 13 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), n < m, n < n0, max(m, n) < n0, n0 ≤ m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
m ≠ n0, max(m, n) < m, n0 < m, n < m, n < n0, max(m, n) < n0, n + 1 < m, max(m, n) + 1 < m ⊦

Rule argument

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; ⊦ m +1 = m + 1; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)

Proof step 14 from proof max-a (previous step) (back to tree)

m ≠ n0, max(m, n) < m, n0 < m, n < m, n < n0, max(m, n) < n0, n + 1 < m, max(m, n) + 1 < m ⊦

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ ¬ n < n; (s) ⊦ m < n → ¬ n < m; (s) Used formulas: m < n

Proof step 15 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), n < m, n < n0, n0 ≤ m, ¬ max(m, n) < n0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m ≠ max(m, n), n < n0, n < m, n < max(m, n), n0 ≤ m, n0 ≤ max(m, n) ⊦

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 16 from proof max-a (previous step) (back to tree)

m ≠ max(m, n), n < n0, n < m, n < max(m, n), n0 ≤ m, n0 ≤ max(m, n) ⊦

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ m < n → (n ≤ m ↔ false); (s) Used formulas: m < n

Proof step 17 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), max(n, n0) ≤ m, ¬ n < n0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
m ≠ max(max(m, n), n0), n ≤ m, n0 ≤ n, n0 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 18 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), n ≤ m, n0 ≤ n, n0 ≤ m ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 19:
m ≠ max(max(m, n), n0), max(m, n) < n0, n0 ≤ m, n0 ≤ n, n ≤ m ⊦
Proof step 21:
m ≠ max(max(m, n), n0), n0 ≤ m, n0 ≤ n, n ≤ m, ¬ max(m, n) < n0 ⊦

Rule argument

fma: max(m, n) < n0

Additional information

Used formulas: max(m, n) < n0

Proof step 19 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), max(m, n) < n0, n0 ≤ m, n0 ≤ n, n ≤ m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
m ≠ n0, max(m, n) + 1 < m, max(m, n) < n0, n0 < m, max(m, n) < m, max(m, n) < n, n ≤ m, n0 ≤ n ⊦

Rule argument

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m +1 = m + 1; (s) trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)

Proof step 20 from proof max-a (previous step) (back to tree)

m ≠ n0, max(m, n) + 1 < m, max(m, n) < n0, n0 < m, max(m, n) < m, max(m, n) < n, n ≤ m, n0 ≤ n ⊦

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → (n ≤ m ↔ false); (s) Used formulas: m < n

Proof step 21 from proof max-a (previous step) (back to tree)

m ≠ max(max(m, n), n0), n0 ≤ m, n0 ≤ n, n ≤ m, ¬ max(m, n) < n0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
m ≠ max(m, n), n ≤ m, n0 ≤ n, n0 ≤ m, n0 ≤ max(m, n) ⊦

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 22 from proof max-a (previous step) (back to tree)

m ≠ max(m, n), n ≤ m, n0 ≤ n, n0 ≤ m, n0 ≤ max(m, n) ⊦

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n; Max-right: ⊦ ¬ m < n → max(m, n) = m; ⊦ m < n → (n ≤ m ↔ false); (s) Used formulas: m < n

Proof Tree for max-a (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.