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Proof step 102 from proof hasSharing-dd-help (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 101:
# sks = 2 + # refs0 + # refs2, # sks0 = 2 + # refs + # refs1, st[r -- sks] = refval(r0), st[r -- sks0] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[# sks0], 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), fits(r, sks0, st), ∀ n. n ≠ 0 ∧ ¬ # sks0 < n → st[r -- firstn(n, sks0)] .refval = ((r ' + refs) + r1 ' + refs1 + r0 ')[n], ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = ((r ' + refs0) + r2 ' + refs2 + r0 ')[n] ⊦ refval(r0) = st[r -- sks0], sks = sks0, r2 = r1, r1 = r0, r0 = jvmref, sks0 = [], sks = [], path∃(str, r2, r1, st), path∃(str, r1, r2, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ firstn(# sks, sks) = sks], [CSIMPSEQ : ⊦ ¬ n < n], [CSIMPSEQ : ⊦ # sks = 0 ↔ sks = []])]]