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Proof step 89 from proof hasSharing-dd-help (previous step)

# refs + # refs1 = # refs2 + # refs0, # sks0 = 2 + # refs2 + # refs0, # refs0 < # refs, fits(r, sks0, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), ∀ n. n ≠ 0 ∧ ¬ # sks0 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[n], 1 + # refs0 ≠ 0 ∧ ¬ # sks0 < 1 + # refs0 → st[r -- firstn(1 + # refs0, sks0)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[1 + # refs0] ⊦ r0 = jvmref, r1 = r0, r2 = r1, # refs = # refs0, refs = [], hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 88:
# refs + # refs1 = # refs2 + # refs0, # sks0 = 2 + # refs2 + # refs0, # refs0 < # refs, fits(r, sks0, st), validrefs(str, (r ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), validrefs(str, (r ' + refs) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), ∀ n. n ≠ 0 ∧ ¬ # sks0 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[n], ¬ # sks0 ≤ # refs0 → st[r -- firstn(1 + # refs0, sks0)] .refval = r2 ⊦ r0 = jvmref, r1 = r0, r2 = r1, # refs = # refs0, refs = [], hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st)

Rule argument

[emptyarg]

Additional information

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