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Proof step 8 from proof validvals-append (previous step)

∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ (rk × val ∈ rkvs → validval(str, rk, val, st)) ∧ (rk0 × val0 ∈ rkvs0 → validval(str, rk0, val0, st))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ rk × val ∈ rkvs → validval(str, rk, val, st)
Proof step 7:
∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ rk0 × val0 ∈ rkvs0 → validval(str, rk0, val0, st)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]