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Proof step 6 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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
∀ n. n→i(m0) < n→i(n) ↔ ¬ n ≤ m0 ⊦ n→i(m0) + 1 < n→i(n) ↔ ¬ n ≤ m0 + 1, n = 0

Rule argument

fma: n = 0

Additional information

Used simplifier rules: nat2int-06: ⊦ ¬ n→i(n) + 1 < 0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 ≤ m; (s) Used formulas: n = 0