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Proof step 10 from proof nat2int-04 (previous step)

n = m +1, n→i(m0) < n→i(m) ↔ ¬ m ≤ m0 ⊦ n = 0, n→i(m0) + 1 < n→i(n) ↔ ¬ n ≤ m0 + 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2