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Proof step 7 from proof nat2int-04 (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 8:
n→i(m0) < n→i(n -1) ↔ ¬ n -1 ≤ m0 ⊦ n = 0, n→i(m0) + 1 < n→i(n) ↔ ¬ n ≤ m0 + 1

Rule argument

pos: left 1 terms: n -1 abort: F discard: T computed: F

Additional information

Used terms: n -1