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Proof step 6 from proof eqtwo-01 (previous step)

∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ++ b ⊦ a = c, b = c

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 7:
c ∈ ∅ ++ a ++ b ↔ c ∈ ∅ ++ c ++ b, ∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ++ b ⊦ a = c, b = c

Rule argument

pos: left 1 terms: c

Additional information

Used terms: c