Commutative Property Variation 1
Substitution 2
Substitution 6
Zero Plus a
Addi to Zero
Transitive Property of Equality Variation 2
Addi Zero Insn 0
Transitive Property Application 2
Pc 0
Addi Zero Insn 1
Pc 1
Byte 3 Stays the Same 1
Addi Zero Insn 2
Pc 2
Byte 3 Stays the Same 2
Byte 4 Stays the Same 2
Addi Insn 3
Pc 3
Byte 3 Stays the Same 3
Byte 4 Stays the Same 3
Addi Insn 4
Pc 4
Byte 3 Stays the Same 4
Byte 5 Stays the Same 4
Truth Propagation Property 2
Beq No Branch 5
Byte 3 Stays the Same 5
Byte 4 Stays the Same 5
Byte 5 Stays the Same 5
Jump Help 6
Byte 3 Stays the Same 6
Byte 4 Stays the Same 6
Byte 5 Stays the Same 6
Addi Insn 7
Pc 7
Byte 3 Stays the Same 7
Byte 4 Stays the Same 7
Addi Insn 8
Pc 8
Byte 3 Stays the Same 8
Byte 5 Stays the Same 8
Beq No Branch 9
Byte 3 Stays the Same 9
Byte 4 Stays the Same 9
Byte 5 Stays the Same 9
Jump Help 10
Byte 3 Stays the Same 10
Byte 4 Stays the Same 10
Byte 5 Stays the Same 10
Addi Insn 11
Pc 11
Byte 3 Stays the Same 11
Byte 4 Stays the Same 11
Addi Insn 12
Pc 12
Byte 3 Stays the Same 12
Byte 5 Stays the Same 12
Transitive Property of Equality Variation 1
Beq Branch
Byte 3 Stays the Same 13
Byte 4 Stays the Same 13
Byte 5 Stays the Same 13
Loop

Proof: Beq Branch

Let's prove the following theorem:

if the following are true:
  • instruction #5 is beq left=3 right=4 imm=1
  • the PC at time 13 = 5
  • value of cell 3 at time 13 = 3
  • value of cell 4 at time 13 = 3

then the PC at time 14 = 7

Proof:

View as a tree | View dependent proofs | Try proving it

Given
1 instruction #5 is beq left=3 right=4 imm=1
2 the PC at time 13 = 5
3 value of cell 3 at time 13 = 3
4 value of cell 4 at time 13 = 3
Proof Table
# Claim Reason
1 value of cell 3 at time 13 = value of cell 4 at time 13 if value of cell 3 at time 13 = 3 and value of cell 4 at time 13 = 3, then value of cell 3 at time 13 = value of cell 4 at time 13
2 the PC at time (13 + 1) = (5 + 1) + 1 if instruction #5 is beq left=3 right=4 imm=1 and the PC at time 13 = 5 and value of cell 3 at time 13 = value of cell 4 at time 13, then the PC at time (13 + 1) = (5 + 1) + 1
3 (5 + 1) + 1 = 7 (5 + 1) + 1 = 7
4 the PC at time (13 + 1) = 7 if the PC at time (13 + 1) = (5 + 1) + 1 and (5 + 1) + 1 = 7, then the PC at time (13 + 1) = 7
5 13 + 1 = 14 13 + 1 = 14
6 the PC at time (13 + 1) = the PC at time 14 if 13 + 1 = 14, then the PC at time (13 + 1) = the PC at time 14
7 the PC at time 14 = 7 if the PC at time (13 + 1) = the PC at time 14 and the PC at time (13 + 1) = 7, then the PC at time 14 = 7
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