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Proof step 10 from proof coprimes-prime (previous step)

prime(i), 0 < i ⊦ filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i - 1)) + (i = 1 ⊃ 1 ';[]) = count-i2j(1, i - 1)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 11:
i ≠ 1, prime(i), 0 < i ⊦ filterFun(λ j. gcd(i, j) = 1, count-i2j(1, i - 1)) + [] = count-i2j(1, i - 1)

Rule argument

pos: right 1

Additional information

Used simplifier rules: counti2j-empty: ⊦ j < i → count-i2j(i, j) = [];%filterFun-base: ⊦ filterFun(intboolfun, []) = [];%gcd-one-01: ⊦ gcd(i, 1) = 1; from specification int-gcd%gcd-com: ⊦ gcd(i, j) = gcd(j, i); from specification int-gcd%Nil: ⊦ [] + ints = ints; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%e: ⊦ i ' ≠ []; from specification list[intlista]%prime-one: ⊦ i ≤ 1 → ¬ prime(i); from specification int-gcd