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Proof step 4 from proof infis-notin-01 (previous step)

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

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:
∀ 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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ∈ (sk0 × val) ' + fis ↔ sk = sk0 ∨ sk ∈ fis], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ ¬ ¬ (rk ∈ st ∧ st[rk] = val)], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0], [CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val])]]