[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 19 from proof equal-minus-one (previous step)

abs(i ^ (j - 1)) = 1, i + i ^ (j - 1) = 0, 0 < j ⊦ i = 0, i = ~1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
j = i0 + 1, abs(i ^ i0) = 1, i + i ^ i0 = 0, 0 < j ⊦ i = 0, i = ~1

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i0] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i0