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Proof step 6 from proof validvals-put-any (previous step)

validval(str, rk, val, st), ∀ rkv. rkv ∈ rkvs → validval(str, rkv .1, rkv .2, st), rk0 × val0 ∈ rkvs ⊦ validval(str, rk0, val0, st[rk, val])

User applied rule all left

And obtained the following new goals

Proof step 5:
validval(str, rk, val, st), rk0 × val0 ∈ rkvs → validval(str, (rk0 × val0) .1, (rk0 × val0) .2, st), rk0 × val0 ∈ rkvs ⊦ validval(str, rk0, val0, st[rk, val])

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (rk0 × val0) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk0 × val0)]]