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Proof step 2 from proof at-other-50 (previous step)

str ≠ str0 ⊦ (st[r -- mkfs(class, ty, str .field) ', val][r0 -- mkfs(class0, ty0, str0 .field) ']) = st[r0 -- mkfs(class0, ty0, str0 .field) ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ stringvar .field = stringvar0 .field ↔ stringvar = stringvar0], [CSIMPSEQ : ⊦ mkfs(class1, ty, fieldvar) = mkfs(class2, ty0, fieldvar0) ↔ ¬ ¬ ( class1 = class2 ∧ ty = ty0 ∧ fieldvar = fieldvar0)], [CSIMPSEQ : ⊦ fs ' = fs0 ' ↔ fs = fs0], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])]]