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Proof step 21 from proof coprimes-min (previous step)

0 < 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))) < 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 insert lemma

And obtained the following new goals

Proof step 22:
⊦ filterFun(intboolfun, ints) ≠ [] → min(ints) ≤ min(filterFun(intboolfun, ints))
Proof step 23:
filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) ≠ [] → min(count-i2j(1, i)) ≤ min(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))), 0 < 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))) < 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

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

Additional information

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