Why does gcc use btq in conjunction with btcq when conditionally set a bit in a uint64_t * memory region

Basically I am trying to understand the code at: https://gcc.godbolt.org/z/7xxb3G

void __attribute__((noinline))
cond_unset_bit(uint64_t * v, uint32_t b) {
    if(__builtin_expect(!!(*v & ((1UL) << b)), 1)) {
        *v ^= ((1UL) << b);
    }
}

compiles to:

cond_unset_bit(unsigned long*, unsigned int):
        movq    (%rdi), %rax
        btq     %rsi, %rax
        jnc     .L6
        btcq    %rsi, %rax
        movq    %rax, (%rdi)
.L6:
        ret

Based on Agner Fog's Instruction Table (skylake is pg 238) btq and btcq have the exact same cost when operating on a register. btcq will also set the carry flag to the previous bit so it seems the exact same logic (with better performance) could be accomplished w/o the btq instruction i.e:

cond_unset_bit(unsigned long*, unsigned int):
        movq    (%rdi), %rax
        btcq    %rsi, %rax
        jnc     .L6
        movq    %rax, (%rdi)
.L6:
        ret

What is the reason for including the btq?

I am tuning for x86_64 / intel skylake chip

Edit: Thanks @Peter Cordes (and for the help on all my other posts :)