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Proof step 18 from proof abs-paths-eq (previous step)

tree[p3 + p].dir?, tree[p3 + p2].dir?, abs(tree, ino0)(dirs), ispath(p3 + p2, ino0, ino1, dirs, files), ispath(p3 + p, ino0, ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, p2 ∈ tree[p3], p ∈ tree[p3], p2 ≠ ε ∧ p ≠ ε → p2.head ≠ p.head, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p2 = p

User applied rule cut formula

And obtained the following new goals

Proof step 19:
tree[p3 + p2].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p3 + p, ino0, ino1, dirs, files), ispath(p3 + p2, ino0, ino1, dirs, files), p ∈ tree[p3], p2 ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, p2 ≠ ε ∧ p ≠ ε → p2.head ≠ p.head, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p = ε, p2 = p, p2 = ε
Proof step 72:
tree[p3 + p2].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p3 + p, ino0, ino1, dirs, files), ispath(p3 + p2, ino0, ino1, dirs, files), p ∈ tree[p3], p2 ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, p2 ≠ ε ∧ p ≠ ε → p2.head ≠ p.head, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p2 = p, p2 ≠ ε ∧ p ≠ ε

Rule argument

fma: p2 ≠ ε ∧ p ≠ ε

Additional information

Used formulas: p2 ≠ ε ∧ p ≠ ε