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Proof step 10 from proof pathex-put-last (previous step)

refs0 = refs1 + r0 ', r ' + refs0 = refs + r0 ', r ∈ refs, validrefs(str, r ' + refs0, st[r -- _type] .type, st[r0 -- sk, val]) ⊦ refs0 = [], sk = _type, r = r0, path∃(str, r, r0, st), junkref(str, r, st), dups(refs0), r ∈ refs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
validrefs(str, r ' + refs1 + r0 ', st[r -- _type] .type, st[r0 -- sk, val]) ⊦ r = r0, sk = _type, r ∈ refs1, dups(refs1), r0 ∈ refs1, junkref(str, r, st), path∃(str, r, r0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ refs → (r ∈ r0 ' + refs ↔ r = r0)], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0], [CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ dups(refs + r ') ↔ r ∈ refs ∨ dups(refs)], [CSIMPSEQ : ⊦ refs + r ' ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ refs + refs0 = refs1 + refs0 ↔ refs = refs1])]]