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Proof step 4 from proof two-mone (previous step)

i = i0 +1, 1 < i, 2 ^ i0 ≤ 0, 2 ≤ i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 ^ i0 < i0, 0 < i0, 2 ^ i0 ≤ 0 ⊦

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ i +1 = i + 1; (s)