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Proof step 8 from proof add-eq2pot-01 (previous step)

i + 1 = i2 + i0, 2 ^ i2 = i1 + 1, 0 ≤ i0, 0 ≤ j, 0 ≤ i ⊦ 2 ^ (i + 1) = 2 ^ i0 + j ↔ j = 2 ^ i0 * i1 ∧ i0 ≤ i2 + i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ 0 ≤ j ∧ 0 ≤ i ∧ 0 ≤ i0 → (2 ^ i = 2 ^ i0 + j ↔ j = 2 ^ i0 * (2 ^ (i - i0) - 1) ∧ i0 ≤ i)

Rule argument

spec: <toplevel> name: add-eq2pot seq: ⊦ 0 ≤ j ∧ 0 ≤ i ∧ 0 ≤ i0 → ( 2 ^ i = 2 ^ i0 + j ↔ j = 2 ^ i0 * (2 ^ (i - i0) - 1) ∧ i0 ≤ i) subst: [j, i0, i] → [j, i0, i + 1] flags: <all paths>

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i ≤ i + j ↔ 0 ≤ j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ i + j - i = j; (s)% ⊦ (* i) + i0 - (* j) = (i - j) + i0; (s)% ⊦ i + 0 = i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs Used terms: j, i0, i + 1