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Proof step 6 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))) ≤ max(count-i2j(1, i)), 0 < i, 1 < i, i ≤ max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))), 1 < max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))), 0 < max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 7:
max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ≤ max(count-i2j(1, i)), 0 < i, 1 < i, i ≤ max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))), 1 < max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))), 0 < max(filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i))) ⊦

Rule argument

pos: left 1

Additional information

Used simplifier rules: max-empty: ⊦ max([]) = 0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs