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Proof step 4 from proof expath-split (previous step)

⊦ (∃ m. ispath(p + q, ino, m, dirs, files)) ↔ (∃ m. ispath(p, ino, m, dirs, files) ∧ ispath(q, m, dirs, files))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ m. ispath(p + q, ino, m, dirs, files) ⊦ ∃ m. ispath(p, ino, m, dirs, files) ∧ ispath(q, m, dirs, files)
Proof step 10:
∃ m. ispath(p, ino, m, dirs, files) ∧ ispath(q, m, dirs, files) ⊦ ∃ m. ispath(p + q, ino, m, dirs, files)

Rule argument

pos: right 1

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