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Proof step 13 from proof coprimes-max (previous step)

filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)) ≠ [] → max(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)) → i0 ≤ max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)))), max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) = i, 1 < i, 0 < i ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 14:
max(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)) → i0 ≤ max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i)))), max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) = i, 1 < i, 0 < i ⊦

Rule argument

pos: left 1

Additional information

Used simplifier rules: max-empty: ⊦ max([]) = 0;%counti2j-empty: ⊦ j < i → count-i2j(i, j) = [];%filterFun-base: ⊦ filterFun(intboolfun, []) = [];%gcd-zero-04: ⊦ gcd(i, 0) = abs(i); from specification int-gcd%gcd-com: ⊦ gcd(i, j) = gcd(j, i); from specification int-gcd