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Proof step 3 from proof infis-notin-01 (previous step)

∀ st, r, sk, val. (fis0 ∈ st[r -- sk, val] ↔ fis0 ∈ st) ∨ sk ∈ fis0 ⊦ sk0 = sk, jvmref -- sk ∈ st[r -- sk0, val0] ∧ fis0 ∈ st[r -- sk0, val0] ∧ st[jvmref -- sk] = val ↔ jvmref -- sk ∈ st ∧ st[jvmref -- sk] = val ∧ fis0 ∈ st, sk0 ∈ fis0

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (st, r, sk0, val0) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])], [EXTRATERMS : (st, r, sk0, val0)]]