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Proof step 11 from proof lenequal (previous step)

# ax0 = # ay0 + # az, unique_al(ay0 + (a × d)' + az), unique_al(ax0), ∀ ay0, b0. # ax0 = # ay0 ∧ unique_al(ay0) ∧ b0 ∈ ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) → b0 ∈ ax0, ∀ a0. a0 ∈ ax0 → a0 ∈ ay0 ∨ a0 ∈ az, b ∈ ay0 ∨ b ∈ az ⊦ b = a, b ∈ ax0, a ∈ ax0, a ∈ ay0, a ∈ az

User applied rule all left

And obtained the following new goals

Proof step 12:
# ax0 = # ay0 + # az, unique_al(ax0), unique_al(ay0 + (a × d)' + az), # ax0 = #(ay0 + az) ∧ unique_al(ay0 + az) ∧ b ∈ ay0 + az ∧ (∀ a. a ∈ ax0 → a ∈ ay0 + az) → b ∈ ax0, ∀ a0. a0 ∈ ax0 → a0 ∈ ay0 ∨ a0 ∈ az, b ∈ ay0 ∨ b ∈ az ⊦ b = a, a ∈ az, a ∈ ay0, a ∈ ax0, b ∈ ax0

Rule argument

pos: left 4 terms: ay0 + az, b abort: F discard: T computed: F

Additional information

Used terms: ay0 + az, b