What are the consequences of alternative type-theoretic definition of homotopy equivalence?

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The standard definition of homotopy equivalence in HoTT is a quadruple of:

  • $f: A \rightarrow B$
  • $g: B \rightarrow A$
  • $p: \operatorname{id} A = g \circ f$
  • $q: \operatorname{id} B = f \circ g$

... where the paths are homogeneous in $A \rightarrow A$ and $B \rightarrow B$ respectively.

My question is what happens if we let the paths be heterogeneous (or in the HoTT book terminology dependent) over some paths $P: (A \rightarrow A) = (A \rightarrow A)$ and $Q: (B \rightarrow B) = (B \rightarrow B)$?

Will it have practical implications for the theory (in comparison to HoTT or CTT), or will the difference be limited to the corresponding categorical semantics?

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Your definition might be an interesting thing, but it's not a definition of anything that most people would call an "equivalence". You're basically saying consider two types $A$ and $B$ and functions $f:A\to B$ and $g:B\to A$ such that there is an automorphism of the type $A\to A$ that sends $g\circ f$ to $\rm id_A$ and an automorphism of the type $B\to B$ that sends $f\circ g$ to $\rm id_B$.

This situation holds, for instance, when $A = \mathbf{1}$ is the unit type and $B = \mathbf{1}+\mathbf{1} = \mathsf{Bool}$ is the two-element type. In this case $B\to B$ is a finite 4-element type, which is "transitive" in that it has an automorphism sending any element to any other element — and of course $A\to A$ is also (equivalent to) a contractible type. Most people would not consider these two types to be equivalent. (-: