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Proof step 7 from proof putar-in (previous step)

∀ 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 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ putArray(val ' + vals, r, i, st) = (putArray(vals, r, i + 1, st)[r -- i ', val])
Proof step 6:
∀ r1, r0, i0, st. r1 ∈ putArray(vals0, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals0 ≠ [] ⊦ r1 ∈ (putArray(vals0, r0, i0 + 1, st)[r0 -- i0 ', val]) ↔ r1 = r0 ∨ r1 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ val ' ≠ []], [CSIMPSEQ : ⊦ ¬ vals + vals0 ≠ [] ↔ vals = [] ∧ vals0 = []], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]