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Proof step 9 from proof i2n-divtwo (previous step)

2 * (n→i(m0) + i) + n→i(n0) = 0, n0 < 2, i < 0 ⊦ m0 = i→n(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m ≠ 1 → (m < 2 ↔ m = 0)
Proof step 11:
n0 ≠ 1, 2 * (n→i(m0) + i) + n→i(n0) = 0, n0 = 0, i < 0 ⊦ m0 = i→n(i)

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: numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%eq-one-neg: ⊦ i * j = ~1 ↔ abs(i) = 1 ∧ i = ~ j; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: n0