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Proof step 41 from proof size-two (previous step)

a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, ∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, b = a ∨ b ∈ s0, a0 = a ∨ a0 ∈ s0 ⊦ a2 ∈ s0 ↔ a2 = a0, a1 ∈ s0 ↔ a1 = b, 1 = # s0, a0 = b, a ∈ s0, ∃ a. s0 = ∅ ++ a

User applied rule all left

And obtained the following new goals

Proof step 42:
a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a2 = a ∨ a2 ∈ s0 ↔ a2 = b ∨ a2 = a0, ∀ a1. a1 = a ∨ a1 ∈ s0 ↔ a1 = b ∨ a1 = a0, a0 = a ∨ a0 ∈ s0, b = a ∨ b ∈ s0 ⊦ a1 ∈ s0 ↔ a1 = b, a2 ∈ s0 ↔ a2 = a0, 1 = # s0, a0 = b, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

pos: left 2 terms: a2 abort: F discard: F computed: T

Additional information

Used terms: a2