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Proof step 2 from proof long-is-cast-01 (previous step)

l2i(i2l(i)) = l2i(i2l(cast(64, i))) ⊦ i2l(i) = i2l(cast(64, i))

User applied rule cut formula

And obtained the following new goals

Proof step 3:
l2i(i2l(i)) = l2i(i2l(cast(64, i))) ⊦ i2l(l2i(i2l(i))) = i2l(l2i(i2l(cast(64, i)))), i2l(i) = i2l(cast(64, i))

Rule argument

fma: i2l(l2i(i2l(i))) = i2l(l2i(i2l(cast(64, i))))

Additional information

Used simplifier rules: ili: ⊦ i2l(l2i(lo)) = lo; Used formulas: i2l(l2i(i2l(i))) = i2l(l2i(i2l(cast(64, i))))