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Proof step 6 from proof eq-one-neg (previous step)

~ i * j = 1, i * j = ~1 ⊦ abs(j) = 1 ∧ j + i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j

Rule argument

spec: int-mult inst: name: eq-one seq: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j subst: [j, i] → [j, ~ i] flags: rewrite, current

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%eq-one: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j;%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs Used terms: j, ~ i