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Proof step 16 from proof coprimes-min (previous step)

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, ∀ 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 17:
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

Rule argument

Additional information

Used simplifier rules: lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs