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Proof step 17 from proof coprimes-min (previous step)

0 < 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, 1 ∈ filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) → min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ≤ 0 ⊦ 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 18:
0 < 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 ⊦ 1 ∈ 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 20:
min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ≤ 0, 0 < 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

Rule argument

pos: left 3

Proof step 17 from proof coprimes-min (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information