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Proof step 5 from proof two-zero-01-01 (previous step)

n0 < 2 ⊦ m0 = 0 ↔ m0 * 2 + n0 = 0 ∨ m0 * 2 + n0 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-basic1 inst: name: twols-01 seq: ⊦ m ≠ 1 → (m < 2 ↔ m = 0) subst: [m] → [n0] flags: rewrite

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%two-one-01: ⊦ 2 * m ≠ 1; from specification nat-mult%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2 Used terms: n0