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Proof step 4 from proof square-ls (previous step)

∀ n0. m0 * m0 + m0 < n0 * n0 + n0 ↔ m0 < n0 ⊦ 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n, n = 0

User applied rule all left

And obtained the following new goals

Proof step 5:
m0 * m0 + m0 < n -1 * n -1 + n -1 ↔ m0 < n -1 ⊦ 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n, n = 0

Rule argument

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

Additional information

Used terms: n -1