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Proof step 9 from proof infis-jvmref (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (sk × val) ' + fis ∈ st ↔ (jvmref -- sk) × val ∈ st ∧ fis ∈ st
Proof step 3:
⊦ (sk × val) ' + fis ∈ st ↔ ¬ ¬ ((jvmref -- sk) × val ∈ st ∧ fis ∈ st)
Proof step 8:
∀ r, st, sk, val. r = jvmref ∨ (fis0 ∈ st[r -- sk, val] ↔ fis0 ∈ st) ⊦ r = jvmref, fis0 ∈ st[r -- sk0, val0] ∧ jvmref -- sk ∈ st ∧ st[jvmref -- sk] = val ↔ jvmref -- sk ∈ st ∧ st[jvmref -- sk] = val ∧ fis0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk × val ∈ st ↔ ¬ ¬ (rk ∈ st ∧ st[rk] = val)], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ r ≠ r0 → (st[r0 -- sk0, val][r -- sk]) = st[r -- sk]], [CSIMPSEQ : ⊦ r ≠ r0 → ( r -- sk ∈ st[r0 -- sk0, val] ↔ r -- sk ∈ st)], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val])]]