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Proof step 27 from proof sharing-twopaths (previous step)

∃ sks. fits((r0 ' + refs0) .first, sks, st) ∧ 1 + # sks = # (r0 ' + refs0) ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[(r0 ' + refs0) .first -- firstn(n, sks)] .refval = (r0 ' + refs0)[n]), st[r -- sks0] = refval(refs0 .last), st[r -- sks] = refval(r0), fits(r, sks0, st), fits(r, sks, st), validrefs(str, r0 ' + refs0, mkclasstype(`java.lang.Object' .class), st) ⊦ refs0 = [], r = jvmref, sks = sks0, sks = [], sks0 = [], r0 = jvmref, refs0 .last = jvmref, sks ⊑ sks0, hasSharing(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
st[r -- sks0] = refval(refs0 .last), st[r -- sks] = refval(r0), # sks1 = # refs0, fits(r, sks0, st), fits(r, sks, st), validrefs(str, r0 ' + refs0, mkclasstype(`java.lang.Object' .class), st), fits(r0, sks1, st), ∀ n. n ≠ 0 ∧ ¬ # refs0 < n → st[r0 -- firstn(n, sks1)] .refval = (r0 ' + refs0)[n] ⊦ refs0 = [], r = jvmref, sks = sks0, sks = [], sks0 = [], r0 = jvmref, refs0 .last = jvmref, sks ⊑ sks0, hasSharing(r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0])]]