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Proof step 117 from proof hasSharing-dd-help (previous step)

∃ sks. fits(((r ' + refs0) + r2 ' + refs2 + r0 ') .first, sks, st) ∧ 1 + # sks = # ((r ' + refs0) + r2 ' + refs2 + r0 ') ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[((r ' + refs0) + r2 ' + refs2 + r0 ') .first -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[n]), validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st) ⊦ r2 = r1, r1 = r0, r0 = jvmref, path∃(str, r2, r1, st), path∃(str, r1, r2, st), hasSharing(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 116:
# sks = 2 + # refs0 + # refs2, validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), fits(r, sks, st), ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[n] ⊦ r2 = r1, r1 = r0, r0 = jvmref, path∃(str, r2, r1, st), path∃(str, r1, r2, st), 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 : n < n0 ⊦ m + ((* n)) = m0 + ((* n0)) ↔ m = m0 + (n0 - n)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0])]]