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Proof step 6 from proof coprimes-min (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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

Rule argument

Additional information

Used simplifier rules: min-in: ⊦ min(ints) ∈ ints ↔ ints ≠ [];