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Proof step 16 from proof bytevalues-eq (previous step)

val .byteval ' + bytevalues(vals0) = bytevalues(vals1), val .byteval ∈ bytevalues(vals1), ∀ vals1. bytevalues(vals0) = bytevalues(vals1) ∧ (∀ val. val ∈ vals0 ∨ val ∈ vals1 → is_bytevalue(val)) → vals0 = vals1, ∀ val0. val0 = val ∨ val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0) ⊦ val ' + vals0 = vals1

User applied rule all left

And obtained the following new goals

Proof step 15:
val .byteval ' + bytevalues(vals0) = bytevalues(vals1), val .byteval ∈ bytevalues(vals1), ∀ vals1. bytevalues(vals0) = bytevalues(vals1) ∧ (∀ val. val ∈ vals0 ∨ val ∈ vals1 → is_bytevalue(val)) → vals0 = vals1, val = val ∨ val ∈ vals0 ∨ val ∈ vals1 → is_bytevalue(val), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0) ⊦ val ' + vals0 = vals1

Rule argument

[EXRARG : pos: left 4, [extquanttermlist: (val)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val)]]