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Proof step 5 from proof putar-in (previous step)

∀ r1, r0, i0, st. r1 ∈ putArray(vals0, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals0 ≠ [] ⊦ r1 = r0 ∨ r1 ∈ putArray(vals0, r0, i0 + 1, st) ↔ r1 = r0 ∨ r1 ∈ st

User applied rule all left

And obtained the following new goals

Proof step 4:
r1 ∈ putArray(vals0, r0, i0 + 1, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals0 ≠ [] ⊦ r1 = r0 ∨ r1 ∈ putArray(vals0, r0, i0 + 1, st) ↔ r1 = r0 ∨ r1 ∈ st

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (r1, r0, i0 + 1, st) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r1, r0, i0 + 1, st)]]