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Proof step 2 from proof pot2-pos-01 (previous step)

⊦ 1 < 2 ^ i ↔ 0 < i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%two-leone: ⊦ 2 ^ i ≤ 1 ↔ i < 1;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) 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% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)