joins-two-sides-differing-in-a-datum-a-symbolize-found-off-the-phi-chain.yaml

# SPDX-FileCopyrightText: Copyright (c) 2025 Objectionary.com
# SPDX-License-Identifier: MIT
---
# yamllint disable rule:line-length
# Both sides hold a bool in the method 'kq', the way the 'rewrite' rules bring
# 'Φ.false' into the shape of a computed bool when EO lowers 'number.floor'.
# The datum of the left one sits in the formation bound to 'if', which is no
# φ of anything, so a 'symbolize' walking the φ chain alone left it a datum
# and the sides never joined (#2008). The line turns every datum of the term
# into a fresh symbol, wherever it stands, so '𝑛3' holds 𝜎3 where '𝑛1' held
# 'FF-', the two sides differ in their symbols alone, and the join makes one
# term of them, joining 𝜎3 with 𝜎2 into 𝜎4.
symbolic: |
  - λ: L_fork
    dataize:
      𝛿1: $.c
    morph:
      𝑛1: $.a
      𝑛2: $.b
    symbolize:
      𝑛3: 𝑛1
      𝑛4: 𝑛2
    join:
      on: 𝛿1
      left: 𝑛3
      right: 𝑛4
location: 'Q.y'
hide-rho: false
input: '⟦ y ↦ ⟦ c ↦ ⟦ λ ⤍ 𝜎1 ⟧, a ↦ ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ ⟦ φ ↦ ⟦ Δ ⤍ FF- ⟧ ⟧ ) ⟧, b ↦ ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ ⟦ φ ↦ ⟦ λ ⤍ 𝜎2 ⟧ ⟧ ) ⟧, λ ⤍ L_fork ⟧.φ ⟧'
result: '⟦⟧'
protocol: |
  𝕄(Φ.y):
    𝔼(L_fork):
      𝛿1·1 := 𝔻(𝜎1:λ)  # 𝔻(ξ.c)
      𝑛1·1 := ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ FF-:Δ:φ ) ⟧  # 𝕄(ξ.a)
      𝑛2·1 := ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ 𝜎2:λ:φ ) ⟧  # 𝕄(ξ.b)
      symbolized 𝜎3 := FF-
      𝑛3·1 := ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ 𝜎3:λ:φ ) ⟧  # symbolize(𝑛1)
      𝑛4·1 := ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ 𝜎2:λ:φ ) ⟧  # 𝑛2
      joined 𝜎4 := 𝛿1·1 ? 𝜎3 : 𝜎2
      𝑛·1·1 := ⟦ φ ↦ ⟦⟧, kq ↦ Φ.bool( if ↦ 𝜎4:λ:φ ) ⟧  # 𝑛3, 𝑛4