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Proof step 3 from proof elim-sub (previous step)

n1 < m, ∀ m, n0. n1 ≤ m → (n0 = m - n1 ↔ m = n0 + n1) ⊦ n0 = m - n1 + 1 ↔ m = n0 + n1 + 1, m = 0

User applied rule all left

And obtained the following new goals

Proof step 4:
n1 < m, n1 ≤ m - 1 → (n0 = m - 1 - n1 ↔ m - 1 = n0 + n1) ⊦ n0 = m - n1 + 1 ↔ m = n0 + n1 + 1, m = 0

Rule argument

pos: left 2 terms: m - 1, n0 abort: F discard: T computed: F

Additional information

Used terms: m - 1, n0