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Proof step 5 from proof disjoint-def-pathex (previous step)

validrefs(str, refs0, st[refs0 .first -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r1 ∧ refs .last = refs0 .last ∧ validrefs(str, refs, st[r1 -- _type] .type, st), refs0 .first ∈ refs, r1 ∈ refs, ∀ refs0, refs1. refs0 = [] ∨ refs1 = [] ∨ ¬ refs0 .first ∈ refs ∨ ¬ refs1 .first ∈ refs ∨ refs0 .first = refs1 .first ∨ refs0 .last ≠ refs1 .last ∨ refs0 .last = jvmref ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ⊦ refs0 .last = jvmref, r1 = refs0 .first, refs0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
refs1 .last = refs0 .last, validrefs(str, refs0, st[refs0 .first -- _type] .type, st), validrefs(str, refs1, st[refs1 .first -- _type] .type, st), refs0 .first ∈ refs, refs1 .first ∈ refs, ∀ refs0, refs1. refs0 = [] ∨ refs1 = [] ∨ ¬ refs0 .first ∈ refs ∨ ¬ refs1 .first ∈ refs ∨ refs0 .first = refs1 .first ∨ refs0 .last ≠ refs1 .last ∨ refs0 .last = jvmref ∨ ¬ validrefs(str, refs0, st[refs0 .first -- _type] .type, st) ∨ ¬ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ⊦ refs0 .last = jvmref, refs1 .first = refs0 .first, refs0 = [], refs1 = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]