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Proof step 8 from proof putar-in (conclusion)

⊦ r1 ∈ putArray(vals, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals ≠ []

User applied rule structural induction

And obtained the following new goals

Proof step 1:
⊦ putArray([], r, i, st) = st
Proof step 7:
∀ r1, r0, i0, st. r1 ∈ putArray(vals0, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals0 ≠ [] ⊦ r1 ∈ putArray(val + vals0, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ val + vals0 ≠ []

Rule argument

[VARWITHFMAVARSARG : vals, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : javavalues generated by [], +;])]]