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Proof step 5 from proof nat2int-04 (previous step)

∀ n. n→i(m0) < n→i(n) ↔ ¬ n ≤ m0 ⊦ n→i(m0 +1) < n→i(n) ↔ ¬ n ≤ m0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ n. n→i(m0) < n→i(n) ↔ ¬ n ≤ m0 ⊦ n→i(m0) + 1 < n→i(n) ↔ ¬ n ≤ m0 + 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2