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Proof step 1 from proof tree-ext (conclusion)

⊦ tree0 = tree1 ↔ (∀ p. (p ∈ tree0 ↔ p ∈ tree1) ∧ (p ∈ tree0 → (tree0[p].dir? ↔ tree1[p].dir?) ∧ (tree0[p].dir? ⊃ tree0[p].meta = tree1[p].meta;tree0[p].fid = tree1[p].fid)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ p. (p ∈ tree0 ↔ p ∈ tree1) ∧ (p ∈ tree0 → (tree0[p].dir? ↔ tree1[p].dir?) ∧ (tree0[p].dir? ⊃ tree0[p].meta = tree1[p].meta;tree0[p].fid = tree1[p].fid)) ⊦ tree0 = tree1

Proof step 1 from proof tree-ext (conclusion)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information