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Proof step 8 from proof eqtwo-01 (previous step)

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: b ∈ ∅ ++ a

Additional information

Used simplifier rules: in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b; Used formulas: b ∈ ∅ ++ a