nests-the-firing-an-operand-brought-down.yaml

# SPDX-FileCopyrightText: Copyright (c) 2025 Objectionary.com
# SPDX-License-Identifier: MIT
---
# yamllint disable rule:line-length
# An operand is brought down by a whole run of 𝔻, so a λ function it fires on
# the way sits one level deeper than the firing waiting for it, and the metas
# of the two are told apart by which firing of the entry each belongs to. The
# judgment each firing is commented with says the same thing from the other
# side: the outer one was fired while 𝕄 reduced the program and the inner one
# by the 𝔻 bringing the operand down, under the binding it was bound to (#1306).
symbolic: |
  - λ: L_number_plus
    dataize:
      𝛿1: $.ρ
      𝛿2: $.x
    𝑛: Φ.number( φ ↦ ⟦ λ ⤍ 𝜎 ⟧ )
location: 'Q.@'
input: >
  ⟦ bytes(φ) ↦ ⟦⟧,
    number(φ) ↦ ⟦ plus(ρ, x) ↦ ⟦ λ ⤍ L_number_plus ⟧ ⟧,
    @ ↦ 5.plus( 6.plus( 7 ) ).plus( 8 ) ⟧
protocol: |
  𝕄(Φ.φ):
    𝑛·0·1 := 5
    𝑛·0·2 := 𝑛·0·1.plus( x ↦ 6.plus( 7 ) )
    𝔼(L_number_plus):
      𝔻(Φ.a🌵0):
        𝕄(Φ.a🌵0):
          𝑛·1·1 := 5
          𝑛·1·2 := Φ.bytes( φ ↦ 40-14-00-00-00-00-00-00:Δ )
        𝛿·1·1 := 40-14-00-00-00-00-00-00  # 5.0
      𝛿1·1 := 𝛿·1·1  # 𝔻(ξ.ρ)
      𝔻(Φ.a🌵1):
        𝕄(Φ.a🌵1):
          𝑛·1·3 := 6
          𝑛·1·4 := 𝑛·1·3.plus( x ↦ 7 )
        𝔼(L_number_plus):
          𝔻(Φ.a🌵2):
            𝕄(Φ.a🌵2):
              𝑛·2·1 := 6
              𝑛·2·2 := Φ.bytes( φ ↦ 40-18-00-00-00-00-00-00:Δ )
            𝛿·2·1 := 40-18-00-00-00-00-00-00  # 6.0
          𝛿1·2 := 𝛿·2·1  # 𝔻(ξ.ρ)
          𝔻(Φ.a🌵3):
            𝕄(Φ.a🌵3):
              𝑛·2·3 := 7
              𝑛·2·4 := Φ.bytes( φ ↦ 40-1C-00-00-00-00-00-00:Δ )
            𝛿·2·2 := 40-1C-00-00-00-00-00-00  # 7.0
          𝛿2·2 := 𝛿·2·2  # 𝔻(ξ.x)
          𝑛·2·5 := Φ.number( φ ↦ 𝜎1:λ )  # 𝑛
          𝕄(𝑛·2·5):
            𝑛·2·6 := Φ.number( φ ↦ 𝜎1:λ )
          𝑛·2·7 := 𝑛·2·6  # 𝕄(𝑛·2·5)
      𝛿2·1 := 𝔻(𝜎1:λ)  # 𝔻(ξ.x)
      𝑛·1·5 := Φ.number( φ ↦ 𝜎2:λ )  # 𝑛
      𝕄(𝑛·1·5):
        𝑛·1·6 := Φ.number( φ ↦ 𝜎2:λ )
      𝑛·1·7 := 𝑛·1·6  # 𝕄(𝑛·1·5)
    𝑛·0·3 := 𝑛·1·6.plus( x ↦ 8 )