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Proof step 1 from proof neq-01 (conclusion) (back to tree)

⊦ m0 ≤ m ∧ n + m ≤ n0 → n0 + 1 ≠ n + m0

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof neq-01 (previous step) (back to tree)

n0 + 1 = n + m0, m0 ≤ m, n + m ≤ n0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof neq-01 (previous step) (back to tree)

n0 + 1 = n + m0, n + m ≤ n0, m0 ≤ m ⊦

User applied rule structural induction

And closed the goal!

Rule argument

indvar: n

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); succ-01: ⊦ m + 1 ≤ n ↔ m < n; le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); nat freely generated by 0, +1;; ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ m +1 = m + 1; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ n + 0 = n; (s)

Proof Tree for neq-01 (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.

the proof tree