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Proof step 25 from proof disjoint-def-pathex (previous step)

refs1 = r2 + refs2, refs1 .last = refs0 .last, r2 ∈ refs, refs0 .first ∈ refs, validrefs(str, refs1, st[r2 -- _type] .type, st), validrefs(str, refs0, st[refs0 .first -- _type] .type, st), ∀ r, r0, r1. ¬ r ∈ refs ∨ ¬ r0 ∈ refs ∨ r = r0 ∨ r1 = jvmref ∨ ¬ path∃(str, r, r1, st) ∨ ¬ path∃(str, r0, r1, st) ⊦ refs1 = [], refs0 = [], r2 = refs0 .first, refs0 .last = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
(r2 ' + refs2) .last = refs0 .last, validrefs(str, refs0, st[refs0 .first -- _type] .type, st), r2 ∈ refs, refs0 .first ∈ refs, validrefs(str, r2 ' + refs2, st[r2 -- _type] .type, st), ∀ r, r0, r1. ¬ r ∈ refs ∨ ¬ r0 ∈ refs ∨ r = r0 ∨ r1 = jvmref ∨ ¬ path∃(str, r, r1, st) ∨ ¬ path∃(str, r0, r1, st) ⊦ refs0 = [], r2 = refs0 .first, refs0 .last = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]