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Proof step 1 from proof two-ls-02 (conclusion)

⊦ 2 * (i + j) < 2 * (i + i0) ↔ j < i0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: numls-10: ⊦ 0 < i → ((* i) * j < (* i) * j0 ↔ j < j0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%ls-add-both: ⊦ i < j → i1 + i < i1 + j; 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%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2