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Proof step 2 from proof in-del (previous step)

p ∈ tree ⊦ q ∈ tree -- p ↔ ¬ p ⊑ q ∧ q ∈ tree, p = ε

User applied rule structural induction

And obtained the following new goals

Proof step 3:
str + p0 ∈ tree, ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ str + p0 = ε, q ∈ tree -- (str + p0) ↔ ¬ str + p0 ⊑ q ∧ q ∈ tree

Rule argument

indvar: p

Additional information

Used simplifier rules: path freely generated by ε, +;;