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Proof step 32 from proof size-two (previous step)

∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a0 = a ∨ a0 ∈ s0 ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, s0 = ∅ ++ a0, a ∈ s0, ∃ a. s0 = ∅ ++ a

User applied rule all left

And obtained the following new goals

Proof step 33:
b = a ∨ b ∈ s0 ↔ b = b ∨ b = a0, ∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a0 = a ∨ a0 ∈ s0 ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ a0, s0 = ∅ ++ b, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

pos: left 1 terms: b abort: F discard: F computed: T

Additional information

Used terms: b