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Proof step 11 from proof eq2 (previous step)

2 ^ i -1 = 1, 0 < i ⊦ i = 1

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-basic1 name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [i, i0]

Additional information

Used simplifier rules: eqone-01: ⊦ i ^ j = 1 ∧ 0 < j → abs(i) = 1;% ⊦ i +1 = i + 1; (s)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i = j -1 ↔ j = i +1; (elim) Used terms: i, i0