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Proof step 5 from proof twodiv-01 (previous step)

∀ m0. m0 | m + 1 → m0 = 1 ∨ m0 = m + 1 ⊦ m = 1, m = 0, 2 | m

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: 2 abort: F discard: T computed: F

Additional information

Used simplifier rules: divi-two-03: ⊦ ¬ 2 | n → 2 | n + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat Used terms: 2