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Proof step 11 from proof disjoint-pathex-forward-both (previous step)

disjoint(str, r0 ' + r1 ', st) ∧ path∃(str, r0, r2, st) ∧ r0 ≠ r1 → disjoint(str, r2 ' + r1 ', st), path∃(str, r0, r2, st), path∃(str, r, r1, st), disjoint(str, r ' + r0 ', st), disjoint(str, r1 ' + r0 ', st) ⊦ r = r0, disjoint(str, r1 ' + r2 ', st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
path∃(str, r0, r2, st), path∃(str, r, r1, st), disjoint(str, r ' + r0 ', st), disjoint(str, r1 ' + r0 ', st), disjoint(str, r0 ' + r1 ', st) ∧ r0 ≠ r1 → disjoint(str, r2 ' + r1 ', st) ⊦ r = r0, disjoint(str, r1 ' + r2 ', st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]