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Proof step 3 from proof eq-two (previous step)

2 ^ i = 2 ^ j, 0 ≤ i, 0 ≤ j ⊦ i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used terms: j, i, 2