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Proof step 4 from proof path-delnonmember (previous step)

path(r0 ' + x0, H), ∀ H, r. path(x0, H) → path(x0, H -- r) ∨ r ∈ x0 ⊦ r = r0, path(r0 ' + x0, H -- r), r ∈ x0

User applied rule all left

And obtained the following new goals

Proof step 5:
path(r0 ' + x0, H), path(x0, H) → path(x0, H -- r) ∨ r ∈ x0 ⊦ r = r0, path(r0 ' + x0, H -- r), r ∈ x0

Rule argument

pos: left 2 terms: H, r abort: F discard: T computed: F

Additional information

Used terms: H, r