joins-a-bare-symbol-with-a-branch-holding-one-symbol-when-lenient.yaml

# SPDX-FileCopyrightText: Copyright (c) 2025 Objectionary.com
# SPDX-License-Identifier: MIT
---
# yamllint disable rule:line-length
# The fork of 'bool.and' from #1817: one branch is a bare symbol, the other a
# call of 'Φ.bool', whose φ is code and not a formation, so its φ chain ends in
# no symbol. The call holds one symbol, '𝜎2', so a lenient join answers a copy
# of the call with '𝜎2' replaced by a fresh '𝜎3', which stands for either
# '𝜎1' or '𝜎2'. The join reads nothing outside the two branches.
symbolic: |
  - λ: L_fork
    morph:
      𝑛1: $.a
      𝑛2: $.b
    join:
      𝑛3: [𝑛1, 𝑛2]
    lenient: true
    𝑛: 𝑛3
location: 'Q.y'
input: |
  ⟦
    bool(if) ↦ ⟦ φ ↦ ξ.if( α0 ↦ ⟦ Δ ⤍ FF- ⟧, α1 ↦ ⟦ Δ ⤍ 00- ⟧ ) ⟧,
    y ↦ ⟦
      a ↦ ⟦ λ ⤍ 𝜎1 ⟧,
      b ↦ ⟦ φ ↦ Φ.bool( if ↦ ⟦ λ ⤍ L_fork, left ↦ ∅, right ↦ ∅, φ ↦ ⟦ λ ⤍ 𝜎2 ⟧ ⟧ ) ⟧,
      λ ⤍ L_fork
    ⟧.φ
  ⟧
result: '⟦ if(left, right) ↦ ⟦ λ ⤍ L_fork, φ ↦ 𝜎3:λ ⟧, φ ↦ if( FF-:Δ, 00-:Δ ) ⟧'
protocol: |
  𝕄(Φ.y):
    𝔼(L_fork):
      𝑛1·1 := 𝜎1:λ  # 𝕄(ξ.a)
      𝑛2·1 := Φ.bool( if(left, right) ↦ ⟦ λ ⤍ L_fork, φ ↦ 𝜎2:λ ⟧ ):φ  # 𝕄(ξ.b)
      𝔻(𝜎3:λ) ∈ { 𝔻(𝜎1:λ), 𝔻(𝜎2:λ) }
      𝑛3·1 := Φ.bool( if(left, right) ↦ ⟦ λ ⤍ L_fork, φ ↦ 𝜎3:λ ⟧ ):φ  # [𝑛1, 𝑛2]
      𝑛·1·1 := Φ.bool( if(left, right) ↦ ⟦ λ ⤍ L_fork, φ ↦ 𝜎3:λ ⟧ ):φ  # 𝑛
      𝑛·1·2 := Φ.bool( if(left, right) ↦ ⟦ λ ⤍ L_fork, φ ↦ 𝜎3:λ ⟧ ):φ  # 𝕄(𝑛·1·1)
    𝑛·0·1 := Φ.bool( if(left, right) ↦ ⟦ λ ⤍ L_fork, φ ↦ 𝜎3:λ ⟧ )