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Proof step 2 from proof two-leone (previous step)

⊦ 2 ^ i ≤ 1 ↔ i ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: notone: ⊦ 1 < i ∧ 0 < j → ¬ i ^ j ≤ 1;%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs