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Proof step 1 from proof two-sub (conclusion)

⊦ 2 * (j0 + i0) + j - 2 * j0 + i = 2 * i0 + j - i

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0;%commutativity: ⊦ i * j = j * i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs