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Proof step 2 from proof two (previous step)

∀ a1. a1 = b ∨ a1 = a ↔ a1 = c ∨ a1 = a0 ⊦ a = a0, a = c ∧ b = a0

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 3:
a = b ∨ a = a ↔ a = c ∨ a = a0, ∀ a1. a1 = b ∨ a1 = a ↔ a1 = c ∨ a1 = a0 ⊦ a = a0, a = c ∧ b = a0

Rule argument

pos: left 1 terms: a

Additional information

Used terms: a