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Proof step 7 from proof coprimes-min (previous step)

0 < i, filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) ≠ [] → (∀ i0. i0 ∈ filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) → min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ≤ i0) ⊦ min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) = 1

User applied rule case distinction

And obtained the following new goals

Proof step 8:
0 < i ⊦ filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) ≠ [], min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) = 1
Proof step 15:
∀ i0. i0 ∈ filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) → min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ≤ i0, 0 < i ⊦ min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) = 1

Rule argument

pos: left 2

Proof step 7 from proof coprimes-min (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information