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Proof step 4 from proof coprimes-min (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ ints ≠ [] → min(ints) ∈ ints ∧ (∀ i. i ∈ ints → min(ints) ≤ i)
Proof step 6:
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

Rule argument

spec: inst: name: min-def seq: ⊦ ints ≠ [] → min(ints) ∈ ints ∧ (∀ i. i ∈ ints → min(ints) ≤ i) subst: [ints] → [filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))] flags: current, []

Additional information

Used terms: filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))