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Proof step 3 from proof injectivity (previous step)

i × j = i0 × j0 ⊦ (i × j).1 = (i0 × j0).1 ∧ (i × j).2 = (i0 × j0).2, i = i0 ∧ j = j0

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information