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

0 ≤ i ⊦ 2 < 2 ^ i ↔ 1 < i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: pot2-pos-01: ⊦ 1 < 2 ^ i ↔ 0 < i;%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%two-le-01: ⊦ 0 ≤ i → (2 ^ i ≤ 2 ↔ i ≤ 1);% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (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 ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)