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Proof step 2 from proof linearity-path (previous step)

path(r ' + x, H), path(r ' + y, H) ⊦ x = y, ∃ r0, y0. x = y + r0 ' + y0, ∃ r0, x0. y = x + r0 ' + x0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
path(r ' + [], H), path(r ' + y, H) ⊦ [] = y, ∃ r0, y0. [] = y + r0 ' + y0, ∃ r0, x0. y = [] + r0 ' + x0
Proof step 7:
∀ r, H, y. path(r ' + x0, H) ∧ path(r ' + y, H) → x0 = y ∨ (∃ r0, y0. x0 = y + r0 ' + y0) ∨ (∃ r0, x1. y = x0 + r0 ' + x1), path(r ' + (r0 + x0), H), path(r ' + y, H) ⊦ r0 + x0 = y, ∃ r, y0. r0 + x0 = y + r ' + y0, ∃ r, x1. y = (r0 + x0) + r ' + x1

Rule argument

indvar: x

Additional information

Used simplifier rules: reflist freely generated by [], +;;